{"id":41,"date":"2023-11-08T13:32:02","date_gmt":"2023-11-08T13:32:02","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/read-algebra-of-polynomial-functions\/"},"modified":"2026-02-05T10:28:12","modified_gmt":"2026-02-05T10:28:12","slug":"3-2-the-algebra-of-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/3-2-the-algebra-of-functions\/","title":{"raw":"3.2 The Algebra of Functions","rendered":"3.2 The Algebra of Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Evaluate polynomial functions.<span style=\"background-color: #ffff00;\">\r\n<\/span><\/li>\r\n \t<li>Evaluate a function with a sum or difference as the input such as [latex]f(x+1)[\/latex].<\/li>\r\n \t<li>Evaluate the sum, difference, product, or quotient of two functions at given inputs.<\/li>\r\n \t<li>Find the sum, difference, product, or quotient of two functions and state the domain.<\/li>\r\n<\/ul>\r\n<\/div>\r\nJust as we have performed algebraic operations on polynomials, we can do the same with polynomial functions.\r\n<h2>Evaluate Polynomial Functions<\/h2>\r\nYou can evaluate polynomial functions similar to how you have been evaluating expressions all along. To evaluate an expression for a value of the variable, you substitute the value for the variable <i>every time<\/i> it appears. Then use the order of operations to find the resulting value for the expression.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nGiven the function [latex]f(x)=3x^{2}-2x+1[\/latex]:\r\n\r\n1) Find [latex]f(-1)[\/latex]\r\n\r\n2) Find [latex]f(x+1)[\/latex]\r\n\r\n[reveal-answer q=\"280466\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"280466\"]\r\n\r\n1)\r\n\r\n[latex]\r\n\\begin{align}\r\n\\require{color}f({\\color{Green}{-1}})&amp;=3({\\color{Green}{-1}})^{2}-2({\\color{Green}{-1}})+1 &amp;&amp;\\color{blue}{\\textsf{substitute -1 for each x in the polynomial}}\\\\\r\n&amp;= 3(1)-2(-1)+1 &amp;&amp;\\color{blue}{\\textsf{follow order of operations, evaluate exponents first, then multiply}}\\\\\r\n&amp;= 3+2+1 &amp;&amp;\\color{blue}{\\textsf{add}}\\\\\r\n&amp;= 6\r\n\\end{align}[\/latex]\r\n\r\n2)\r\n\r\n[latex]\r\n\\begin{align}\r\n\\require{color}f({\\color{Green}{x+1}})&amp;=3({\\color{Green}{x+1}})^{2}-2({\\color{Green}{x+1}})+1 &amp;&amp;\\color{blue}{\\textsf{substitute x+1 for each x in the polynomial}}\\\\\r\n&amp;= 3(x+1)(x+1)-2(x+1)+1 &amp;&amp;\\color{blue}{\\textsf{follow order of operations}}\\\\\r\n&amp;= 3(x^2+2x+1)-2x-2+1\\\\ &amp;= 3x^2+6x+3-2x-2+1 &amp;&amp;\\color{blue}{\\textsf{combine like terms}}\\\\\r\n&amp;= 3x^2+4x+2\r\n\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show more examples of evaluating polynomials for given values of the variable.\r\n\r\nhttps:\/\/youtu.be\/Id6UovYjd-M?si=juTNY-ELRx3tHqt\r\n\r\nIn this section, we will focus on how to perform\u00a0algebraic operations on polynomial functions and introduce related notation.\r\n<div class=\"textbox shaded\">\r\n\r\nThe four basic operations on functions are adding, subtracting, multiplying, and dividing. If [latex]f[\/latex] and [latex]g[\/latex] are functions and [latex]x[\/latex] is in the domain of both functions, then:\r\n\r\nAddition: [latex](f + g)(x) = f(x)+ g(x)[\/latex]\r\n\r\nSubtraction: [latex](f \u2212 g)(x)= f(x) \u2212 g(x)[\/latex]\r\n\r\nMultiplication: [latex](f \u00b7 g)(x)= f(x)\\cdot g(x)[\/latex]\r\n\r\nDivision: [latex]\\left(\\dfrac{f}{ g}\\right) (x) = \\dfrac{f(x)}{g(x)}[\/latex], provided [latex]g(x) \\not= 0[\/latex]\r\n\r\nWe will focus on applying these operations to polynomial functions in this section.\r\n\r\n<\/div>\r\nOperations such as addition, subtraction, multiplication, and division can be used to produce a new function from two or more functions. The domain of the new function will be the intersection of the domains of the initial functions.\r\n<h2>Add and Subtract Polynomial Functions<\/h2>\r\nAdding and subtracting polynomial functions is the same as adding and subtracting polynomials. When you evaluate a sum or difference of functions, you can either evaluate first or perform the operation on the functions first as we will see. Our next examples describe the notation used to add and subtract polynomial functions.\r\n<div class=\"textbox exercises\">\r\n<h3>ExAMPLE<\/h3>\r\nLet [latex]f(x)=3x^2-6x-2[\/latex] and [latex]g(x)=4x-1[\/latex]\r\n\r\nCompute the following:\r\n\r\n[latex]1)\\, f(-1)+g(4)\\\\[\/latex]\r\n\r\n[latex]2)\\, f(-1)-g(4)\\\\[\/latex]\r\n\r\n[latex]3)\\, g(-2)-g(3)\\\\[\/latex]\r\n\r\n[reveal-answer q=\"661294\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"661294\"]\r\n\r\n1)\r\n\r\n[latex] f(-1)+g(4)[\/latex]\r\n\r\nTo find [latex]f(-1),[\/latex] substitute [latex]\\require{color}\\color{Green}{-1}[\/latex] in place of [latex]x[\/latex] in the given function.\u00a0To find [latex]f(4)[\/latex] substitute\u00a0[latex]\\require{color}\\color{Green}{4}[\/latex] in place of [latex]x[\/latex] in the given function.\r\n\r\n[latex]\\begin{align}f(\\color{Green}{-1}\\color{black}{)} &amp;=3(\\color{Green}{-1}\\color{black}{)^2-6(}\\color{Green}{-1}\\color{black}{)-2}\\\\ &amp;= (3)(1)-6(-1)-2\\\\ &amp;= 3+6-2\\\\ &amp;= 7 \\end{align}[\/latex]\r\n\r\nand\r\n\r\n[latex]\\begin{align}g(\\color{Green}{4}\\color{black}{)} &amp;=4(\\color{Green}{4}\\color{black}{)-1(}\\\\ &amp;= 16-1\\\\ &amp;=15 \\end{align}[\/latex]\r\n\r\nWe now have, [latex]f(-1)=7[\/latex] and [latex]g(4)=15[\/latex].\r\n\r\nLet's substitute these in place of [latex]f(-1)[\/latex] and [latex]g(4)[\/latex].\r\n\r\n[latex]\\require{color} f(\\color{Green}{-1}\\color{black}{)+g(}\\color{Green}{4}\\color{black}{)}[\/latex]\r\n\r\n[latex]\\begin{align}&amp;= 7+15\\\\ &amp;= 22\\end{align}\\\\\\\\[\/latex]\r\n\r\n2)\r\n\r\n[latex]f(-1)-g(4)\\\\[\/latex]\r\n\r\nIn part 1 we found [latex]f(-1)=7[\/latex] and [latex]g(4)=15[\/latex].\r\n\r\nLet's substitute these in place of [latex]f(-1)[\/latex] and [latex]g(4)[\/latex].\r\n\r\n[latex] f(-1)-g(4)[\/latex]\r\n\r\n[latex]\\begin{align}&amp;= 7-15\\\\ &amp;= -8\\end{align}\\\\\\\\[\/latex]\r\n\r\n3)\r\n\r\n[latex]g(-2)-g(3)\\\\[\/latex]\r\n\r\nTo find [latex]g(-2),[\/latex] substitute [latex]\\require{color}\\color{Green}{-2}[\/latex] in place of [latex]x[\/latex] in the given function.\u00a0To find [latex]g(3)[\/latex] substitute\u00a0[latex]\\require{color}\\color{Green}{3}[\/latex] in place of [latex]x[\/latex] in the given function.\r\n\r\n[latex]\\begin{align}g(\\color{Green}{-2}\\color{black}{)} &amp;=4(\\color{Green}{-2}\\color{black}{)-1}\\\\ &amp;= -8-1\\\\ &amp;= -9 \\end{align}[\/latex]\r\n\r\nand\r\n\r\n[latex]\\begin{align}g(\\color{Green}{3}\\color{black}{)} &amp;=4(\\color{Green}{3}\\color{black}{)-1}\\\\ &amp;= 12-1\\\\ &amp;=11 \\end{align}[\/latex]\r\n\r\nWe now have, [latex]g(-2)=-9[\/latex] and [latex]g(3)=11[\/latex].\r\n\r\nLet's substitute these in place of [latex]g(-2)[\/latex] and [latex]g(3)[\/latex].\r\n\r\n[latex]\\require{color} g(\\color{Green}{-2}\\color{black}{)-g(}\\color{Green}{3}\\color{black}{)}[\/latex]\r\n\r\n[latex]\\begin{align}&amp;= -9-11\\\\ &amp;= -20\\end{align}\\\\\\\\[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nHere is a video with more examples like the one above.\r\n\r\nhttps:\/\/youtu.be\/cqDZl1DrZIg\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nLet [latex]f(x)=2x^3-5x+3[\/latex] and [latex]h(x)=x-5[\/latex],\r\n\r\nFind the following:\r\n\r\n[latex]1) (f+h)(x)[\/latex] and state the domain of the combined function.\r\n\r\n[latex]2) (h-f)(x)[\/latex] and state the domain of the combined function.\r\n[reveal-answer q=\"295585\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"295585\"]\r\n\r\n1)\r\n\r\n[latex]\\require{color}\\begin{align}(f+h)(x)=f(x)+ h(x) &amp; =(2x^3-5x+3)+(x-5) \\\\ &amp; =2x^3-5x+x+3-5 &amp;&amp; \\color{blue}{\\textsf{group like terms}} \\\\ &amp; =2x^3-4x-2 &amp;&amp; \\color{blue}{\\textsf{simplify}}\\end{align}[\/latex]\r\n\r\nThe domain of the combined function is the intersection of the domains of the original functions.\r\nThe domain of [latex]f(x)=2x^3-5x+3[\/latex] is [latex](-\\infty, \\infty)[\/latex]\u00a0because there are no restrictions on the domain.\r\nThe domain of [latex]h(x)=x-5[\/latex] is [latex](-\\infty, \\infty)[\/latex] because there are also no restrictions on the domain.\r\nThe intersection of the two domains will also be\u00a0[latex](-\\infty, \\infty)[\/latex].\r\n\r\n2)\r\n\r\n[latex]\\require{color}\\begin{align}(h-f)(x)=h(x)-f(x) &amp; =(x-5)-(2x^3-5x+3)\\\\ &amp; =x-5-2x^2+5x-3 &amp;&amp; \\color{blue}{\\textsf{distribute negative}} \\\\ &amp; =-2x^2+6x-8 &amp;&amp; \\color{blue}{\\textsf{simplify}}\\end{align}[\/latex]\r\n\r\nThe domain of the combined function is the intersection of the domains of the original functions.\r\nThe domain of [latex]f(x)=2x^3-5x+3[\/latex] is [latex](-\\infty, \\infty)[\/latex]\u00a0because there are no restrictions on the domain.\r\nThe domain of [latex]h(x)=x-5[\/latex] is [latex](-\\infty, \\infty)[\/latex] because there are also no restrictions on the domain.\r\nThe intersection of the two domains will also be\u00a0[latex](-\\infty, \\infty)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our next example, we will evaluate a sum and difference of functions and show that you can get to the same result in one of two ways.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nLet [latex]f(x)=2x^3-5x+3[\/latex] and [latex]h(x)=x-5[\/latex]\r\n\r\nEvaluate:\u00a0[latex](f+h)(2)[\/latex]\r\n\r\nShow that you get the same result by\r\n\r\n1) Evaluating the functions first, then performing the indicated operation on the result.\r\n\r\n2) Performing the operation on the functions first then evaluating the result.\r\n\r\n[reveal-answer q=\"754772\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"754772\"]\r\n\r\n1) First, we will evaluate the functions separately:\r\n\r\n[latex]\\require{color}\\begin{align}f(\\color{Green}{2}\\color{black}{)} &amp;= 2(\\color{Green}{2}\\color{black}{)}^3-5(\\color{Green}{2}\\color{black}{)+3}\\\\ &amp;=16-10+3\\\\ &amp;=9 \\end{align}[\/latex]\r\n\r\n[latex]\\begin{align}h(\\color{Green}{2}\\color{black}{)} &amp;= (\\color{Green}{2}\\color{black}{)-5}\\\\ &amp;=-3 \\end{align}[\/latex]\r\n\r\nNow we will perform the indicated operation using the results:\r\n\r\n[latex]\\begin{align}(f+h)(2) &amp;= f(2)+h(2)\\\\ &amp;= 9+(-3)\\\\ &amp;=6 \\end{align}[\/latex]\r\n\r\n&nbsp;\r\n\r\n2) We can get the same result by adding the functions first and then evaluating the result at\u00a0[latex]x=2[\/latex].\r\n\r\n[latex]\\begin{align} (f+h)(x) &amp;=f(x)+h(x)\\\\ &amp;=(2{x}^{3}-5x+3)+(x-5)\\\\ &amp;=2x^3-4x-2 \\end{align}[\/latex].\r\n\r\nNow we can evaluate this result at\u00a0[latex]x=2[\/latex]\r\n\r\n[latex]\\require{color}\\begin{align} (f+h)(\\color{Green}{2}\\color{black}{)} &amp;=2(\\color{Green}{2}\\color{black}{)^3-4(}\\color{Green}{2}\\color{black}{)-2}\\\\ &amp;=16-8-2\\\\ &amp;=6 \\end{align}[\/latex]\r\n\r\nBoth methods give the same result, and both require about the same amount of work.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>ExAMPLE<\/h3>\r\nGiven: [latex]f(x)=x^2-3x\\\\[\/latex]\r\n\r\nFind:\r\n\r\n[latex]1)f(x+1)-f(2)\\\\[\/latex]\r\n\r\n[latex]2)f(x+1)+f(x)\\\\[\/latex]\r\n\r\n[latex]3)f(x+h)-f(x)\\\\[\/latex]\r\n\r\n[reveal-answer q=\"830895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"830895\"]\r\n\r\n[latex]1)f(x+1)-f(2)\\\\[\/latex]\r\n\r\nTo find [latex]f(x+1),[\/latex] substitute [latex]\\require{color}\\color{Green}{x+1}[\/latex] in place of [latex]x[\/latex] in the given function.\u00a0To find [latex]f(2)[\/latex] substitute\u00a0[latex]\\require{color}\\color{Green}{2}[\/latex] in place of [latex]x[\/latex] in the given function.\r\n\r\n[latex]\\require{color}\\begin{align}f(x) &amp;= x^2-3x\\\\f(\\color{Green}{x+1}\\color{black}{)}&amp;= (\\color{Green}{x+1}\\color{black}{)^{2}-3}(\\color{Green}{x+1}\\color{black}{)}\\\\ f(\\color{Green}{2}\\color{black}{)} &amp;=(\\color{Green}{2}\\color{black}{)}^{2}\\color{black}{-3(}\\color{Green}{2}\\color{black}{)}\\end{align}[\/latex]\r\n\r\nWe now have, [latex]f(x+1)=\\color{BurntOrange}{{(x+1)}^{2}-3(x+1)}[\/latex] and [latex]f(2)=\\color{Plum}{(2)^2-3(2)}[\/latex].\r\n\r\nLet's substitute these in place of [latex]f(x+1)[\/latex] and [latex]f(2)[\/latex].\r\n\r\n[latex]\\require{color}\\color{BurntOrange}{f(x+1)}\\color{black}{+}\\color{Plum}{f(2)}[\/latex]\r\n\r\n[latex]\\require{color}\\begin{align} &amp;=\\color{BurntOrange}{{(x+1)}^{2}-3(x+1)}\\color{black}{+}\\color{Plum}{(2)^2-3(2)} &amp;&amp;\\color{blue}{\\textsf{substitute}}\\\\ &amp;= \\color{BurntOrange}{(x+1)(x+1)-3(x+1)}\\color{black}{+}\\color{Plum}{(2)(2)-3(2)}\\\\ &amp;= \\color{BurntOrange}{x^2+2x+1-3x-3}\\,\\color{black}{+} \\color{Plum}{\\,4-6} &amp;&amp;\\color{blue}{\\textsf{multiply}}\\\\ &amp;= \\color{BurntOrange}{x^2-x-2}\\,\\color{black}{+}\\,\\color{Plum}{(-2)} &amp;&amp;\\color{blue}{\\textsf{combine like terms}}\\\\ &amp;= x^2-x-4 &amp;&amp;\\color{blue}{\\textsf{combine like terms}}\\end{align}\\\\[\/latex]\r\n\r\n&nbsp;\r\n\r\n[latex]2)f(x+1)+f(x)\\\\[\/latex]\r\n\r\nTo find [latex]f(x+1),[\/latex] substitute [latex]\\require{color}\\color{Green}{x+1}[\/latex] in place of [latex]x[\/latex] in the given function.\r\n\r\n[latex]\\require{color}\\begin{align}f(x) &amp;= x^2-3x\\\\f(\\color{Green}{x+1}\\color{black}{)} &amp;= (\\color{Green}{x+1}\\color{black}{)^{2}-3}(\\color{Green}{x+1}\\color{black}{)}\\end{align}[\/latex]\r\n\r\nWe now have, [latex]f(x+1)=\\color{BurntOrange}{{(x+1)}^{2}-3(x+1)}[\/latex] and [latex]f(x)=\\color{Plum}{x^2-3x}[\/latex].\r\n\r\nLet's substitute these in place of [latex]f(x+1)[\/latex] and [latex]f(x)[\/latex].\r\n\r\n[latex]\\require{color}\\color{BurntOrange}{f(x+1)}\\color{black}{+}\\color{Plum}{f(x)}[\/latex]\r\n\r\n[latex]\\require{color}\\begin{align} &amp;=\\color{BurntOrange}{{(x+1)}^{2}-3(x+1)}\\color{black}{+}\\color{Plum}{x^2-3x} &amp;&amp;\\color{blue}{\\textsf{substitute}}\\\\ &amp;= \\color{BurntOrange}{(x+1)(x+1)-3(x+1)}\\color{black}{+}\\color{Plum}{x^2-3x}\\\\ &amp;= \\color{BurntOrange}{x^2+2x+1-3x-3}\\color{black}{+}\\color{Plum}{x^2-3x} &amp;&amp;\\color{blue}{\\textsf{multiply}}\\\\ &amp;= 2x^2-4x-2 &amp;&amp;\\color{blue}{\\textsf{combine like terms}}\\end{align}\\\\[\/latex]\r\n\r\n&nbsp;\r\n\r\n[latex]3)f(x+h)-f(x)\\\\[\/latex]\r\n\r\nTo find [latex]f(x+h),[\/latex] substitute [latex]\\require{color}\\color{Green}{x+h}[\/latex] in place of [latex]x[\/latex] in the given function.\r\n\r\n[latex]\\require{color}\\begin{align}f(x) &amp;= x^2-3x\\\\f(\\color{Green}{x+h}\\color{black}{)} &amp;= (\\color{Green}{x+h}\\color{black}{)^{2}-3}(\\color{Green}{x+h}\\color{black}{)}\\end{align}[\/latex]\r\n\r\nWe now have, [latex]f(x+h)=\\color{BurntOrange}{{(x+h)}^{2}-3}(x+h)[\/latex] and [latex]f(x)=\\color{Plum}{x^2-3x}[\/latex].\r\n\r\nLet's substitute these in place of [latex]f(x+h)[\/latex] and [latex]f(x)[\/latex].\r\n\r\n[latex]\\require{color}\\color{BurntOrange}{f(x+h)}\\color{black}{-}\\color{Plum}{f(x)}[\/latex]\r\n\r\n[latex]\\require{color}\\begin{align} &amp;=\\color{BurntOrange}{{(x+h)}^{2}-3(x+h)}\\color{black}{-}\\color{Plum}{(x^2-3x)} &amp;&amp;\\color{blue}{\\textsf{substitute}}\\\\ &amp;= \\color{BurntOrange}{(x+h)(x+h)-3(x+h)}\\color{black}{-}\\color{Plum}{(x^2-3x)}\\\\ &amp;= \\color{BurntOrange}{x^2+2xh+h^2-3x-3h}\\color{Plum}{-x^2+3x} &amp;&amp;\\color{blue}{\\textsf{distributive property}}\\\\ &amp;= 2xh+h^2-3h &amp;&amp;\\color{blue}{\\textsf{combine like terms}}\\end{align}\\\\[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2>Multiply and Divide Polynomial Functions<\/h2>\r\nTo multiply and divide polynomial functions, we can either evaluate the function first and then perform the indicated operation or vice-versa.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nLet [latex]g(t)=t+7[\/latex] and [latex]f(t)=5t^2-3[\/latex]\r\n\r\n1) Find [latex](g \u00b7 f)(t)[\/latex]\r\n\r\n2) Evaluate [latex](g \u00b7 f)(-1)[\/latex]\r\n\r\n3) Evaluate [latex](f \\cdot g)(-1)[\/latex]\r\n\r\nShould Part 2 and Part 3 have the same answer? Explain.\r\n\r\n[reveal-answer q=\"48983\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"48983\"]\r\n<p style=\"text-align: left;\">1)<\/p>\r\n<p style=\"text-align: left;\">[latex]\\begin{align}\\require{color}(g \u00b7 f)(t) &amp; =(t+7)(5t^2-3) \\\\ &amp; =t\\cdot(5t^2)+t\\cdot(-3)+7\\cdot(5t^2)+7\\cdot(-3) &amp;&amp;\\color{blue}{\\textsf{apply the distributive property}} \\\\ &amp; =5t^3-3t+35t^2-21 &amp;&amp;\\color{blue}{\\textsf{simplify}}\\\\ &amp; =5t^3+35t^2-3t-21 &amp;&amp;\\color{blue}{\\textsf{write in descending order}}\\end{align}\\\\[\/latex]<\/p>\r\n2)\r\n\r\nEvaluate\u00a0[latex](g \u00b7 f)(-1)[\/latex]\r\n<p style=\"text-align: left;\">[latex]\\begin{align}\\require{color}(g \u00b7 f)(t) &amp;=5t^3+35t^2-3t-21 &amp;&amp; \\color{blue}{\\textsf{from above}}\\\\(g \u00b7 f)(-1) &amp;=5(-1)^3+35(-1)^2-3(-1)-21 &amp;&amp; \\color{blue}{\\textsf{evaluate exponents and multiply}}\\\\ &amp;=-5+35+3-21 &amp;&amp; \\color{blue}{\\textsf{add and subtract}}\\\\ &amp;=12\\end{align}[\/latex]<\/p>\r\n3)\r\n\r\nEvaluate [latex](f \\cdot g)(-1)[\/latex]\r\n\r\nWe could take the same steps as we did above by finding\u00a0[latex](f \\cdot g)(t)[\/latex] and then substituting [latex]-1[\/latex] in place of [latex]t[\/latex]. Here is another way you can evaluate\u00a0[latex](f \\cdot g)(-1)[\/latex].\r\n\r\n[latex](f \\cdot g)(-1)= f(-1) \\cdot g(-1)[\/latex]\r\n\r\nLet's first find [latex]f(-1)[\/latex] and [latex]g(-1)[\/latex]. Then multiply the results together.\r\n\r\n[latex]\\begin{align}f(\\color{Green}{-1}\\color{black}{)} &amp;= 5(\\color{Green}{-1}\\color{black}{)^2-3}\\\\ &amp;= 5(1)-3\\\\ &amp;= 5-3\\\\ &amp;= 2 \\end{align}[\/latex]\r\n\r\n[latex]\\begin{align}g(\\color{Green}{-1}\\color{black}{)} &amp;= \\color{Green}{-1}\\color{black}{+7}\\\\ &amp;= 6\\end{align}[\/latex]\r\n\r\nMultiply results together.\r\n\r\n[latex](f \\cdot g)(-1) = f(-1) \\cdot g(-1) = (2)\\cdot (6) = 12[\/latex]\r\n\r\nPart 2 and Part 3 have the same answer because multiplication is commutative.\r\n\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, we will divide polynomial functions and then evaluate the new function.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nGiven [latex]p(x)=4x[\/latex] and [latex]r(x)=x+3[\/latex]\r\n\r\nFind [latex]\\left(\\dfrac{p}{r}\\right)(x)[\/latex] and state its domain. Then evaluate\u00a0[latex]\\left(\\dfrac{p}{r}\\right)(-2)[\/latex]\r\n[reveal-answer q=\"374371\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"374371\"]\r\n\r\n[latex]\\left(\\dfrac{p}{r}\\right)(x)=\\dfrac{p(x)}{r(x)}=\\dfrac{4x}{x+3},\\\\[\/latex]\r\n\r\nDomain: All real numbers except [latex]x\\not=-3.[\/latex]The value [latex]x=-3[\/latex] will make the denominator equal to zero and therefore the expression will be undefined.[latex]\\\\[\/latex]\r\n\r\nDomain in Interval Notation: [latex](-\\infty, -3)\\bigcup(-3, \\infty)\\\\[\/latex]\r\n\r\nDomain in Set-Builder Notation: [latex]\\{ x | x\\not=-3 \\}\\\\\\\\[\/latex]\r\n\r\nEvaluate this quotient for\u00a0[latex]x = -2[\/latex].\r\n\r\n[latex]\\begin{align}\\left(\\dfrac{p}{r}\\right)(-2) &amp;=\\dfrac{p(-2)}{r(-2)}\\\\\\\\ &amp;=\\dfrac{4(-2)}{-2+3}\\\\\\\\ &amp;=\\dfrac{-8}{1}\\\\\\\\ &amp;=-8\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our next example we will find function values from a graph.\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE<\/h3>\r\nFor the functions [latex]\\color{DarkRed}{f(x)}[\/latex], red graph, and [latex]\\color{blue}{g(x)}[\/latex], blue graph, pictured, find the function values.\r\n\r\n[latex]1)f(-1)[\/latex]\r\n\r\n[latex]2)(f-g)(2)[\/latex]\r\n\r\n[latex]3)(f+g)(0)[\/latex]\r\n\r\n[latex]4)\\left(\\dfrac{f}{g}\\right)(2)\\\\[\/latex]\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/read-algebra-of-polynomial-functions\/desmos-graph-31\/\" rel=\"attachment wp-att-1574\"><img class=\"alignnone wp-image-1574\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/desmos-graph-31.png\" alt=\"Two curves are shown: red curve increasing going through (negative 1, negative 2), (0, negative 1), (1,0) and (2,7). Blue curve increasing through (0, negative 5), (1, negative 2), (2,1) and (9,4).\" width=\"454\" height=\"454\" \/><\/a>\r\n\r\n[reveal-answer q=\"597368\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"597368\"]\r\n\r\n1)[latex]\\begin{align}\\require{color}f(-1)=-2 &amp;&amp;\\color{blue}{\\textsf{when the input is -1, the output is -2}} \\end{align}\\\\[\/latex]\r\n\r\n2)[latex]\\begin{align}\\require{color}\\left(f-g \\right)(2) &amp;=f(2)-g(2)\\\\ &amp;=7-1 &amp;&amp;\\color{blue}{\\textsf{subtract}}\\\\ &amp;=6 \\end{align}\\\\[\/latex]\r\n\r\n3)[latex]\\begin{align}\\require{color}\\left(f+g \\right)(0) &amp;=f(0)+g(0)\\\\ &amp;=-1+(-5) &amp;&amp;\\color{blue}{\\textsf{add}}\\\\ &amp;=-6 \\end{align}\\\\[\/latex]\r\n\r\n4)[latex]\\left(\\dfrac{f}{g}\\right)(2)=\\dfrac{f(2)}{g(2)}=\\dfrac{7}{1}=7[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNow try this next example:\r\n<div class=\"textbox tryit\">[ohm_question sameseed=1]288649[\/ohm_question]<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Evaluate polynomial functions.<span style=\"background-color: #ffff00;\"><br \/>\n<\/span><\/li>\n<li>Evaluate a function with a sum or difference as the input such as [latex]f(x+1)[\/latex].<\/li>\n<li>Evaluate the sum, difference, product, or quotient of two functions at given inputs.<\/li>\n<li>Find the sum, difference, product, or quotient of two functions and state the domain.<\/li>\n<\/ul>\n<\/div>\n<p>Just as we have performed algebraic operations on polynomials, we can do the same with polynomial functions.<\/p>\n<h2>Evaluate Polynomial Functions<\/h2>\n<p>You can evaluate polynomial functions similar to how you have been evaluating expressions all along. To evaluate an expression for a value of the variable, you substitute the value for the variable <i>every time<\/i> it appears. Then use the order of operations to find the resulting value for the expression.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Given the function [latex]f(x)=3x^{2}-2x+1[\/latex]:<\/p>\n<p>1) Find [latex]f(-1)[\/latex]<\/p>\n<p>2) Find [latex]f(x+1)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q280466\">Show Solution<\/span><\/p>\n<div id=\"q280466\" class=\"hidden-answer\" style=\"display: none\">\n<p>1)<\/p>\n<p>[latex]\\begin{align}  \\require{color}f({\\color{Green}{-1}})&=3({\\color{Green}{-1}})^{2}-2({\\color{Green}{-1}})+1 &&\\color{blue}{\\textsf{substitute -1 for each x in the polynomial}}\\\\  &= 3(1)-2(-1)+1 &&\\color{blue}{\\textsf{follow order of operations, evaluate exponents first, then multiply}}\\\\  &= 3+2+1 &&\\color{blue}{\\textsf{add}}\\\\  &= 6  \\end{align}[\/latex]<\/p>\n<p>2)<\/p>\n<p>[latex]\\begin{align}  \\require{color}f({\\color{Green}{x+1}})&=3({\\color{Green}{x+1}})^{2}-2({\\color{Green}{x+1}})+1 &&\\color{blue}{\\textsf{substitute x+1 for each x in the polynomial}}\\\\  &= 3(x+1)(x+1)-2(x+1)+1 &&\\color{blue}{\\textsf{follow order of operations}}\\\\  &= 3(x^2+2x+1)-2x-2+1\\\\ &= 3x^2+6x+3-2x-2+1 &&\\color{blue}{\\textsf{combine like terms}}\\\\  &= 3x^2+4x+2  \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we show more examples of evaluating polynomials for given values of the variable.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Evaluating functions given their formula | Functions and their graphs | Algebra II | Khan Academy\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Id6UovYjd-M?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In this section, we will focus on how to perform\u00a0algebraic operations on polynomial functions and introduce related notation.<\/p>\n<div class=\"textbox shaded\">\n<p>The four basic operations on functions are adding, subtracting, multiplying, and dividing. If [latex]f[\/latex] and [latex]g[\/latex] are functions and [latex]x[\/latex] is in the domain of both functions, then:<\/p>\n<p>Addition: [latex](f + g)(x) = f(x)+ g(x)[\/latex]<\/p>\n<p>Subtraction: [latex](f \u2212 g)(x)= f(x) \u2212 g(x)[\/latex]<\/p>\n<p>Multiplication: [latex](f \u00b7 g)(x)= f(x)\\cdot g(x)[\/latex]<\/p>\n<p>Division: [latex]\\left(\\dfrac{f}{ g}\\right) (x) = \\dfrac{f(x)}{g(x)}[\/latex], provided [latex]g(x) \\not= 0[\/latex]<\/p>\n<p>We will focus on applying these operations to polynomial functions in this section.<\/p>\n<\/div>\n<p>Operations such as addition, subtraction, multiplication, and division can be used to produce a new function from two or more functions. The domain of the new function will be the intersection of the domains of the initial functions.<\/p>\n<h2>Add and Subtract Polynomial Functions<\/h2>\n<p>Adding and subtracting polynomial functions is the same as adding and subtracting polynomials. When you evaluate a sum or difference of functions, you can either evaluate first or perform the operation on the functions first as we will see. Our next examples describe the notation used to add and subtract polynomial functions.<\/p>\n<div class=\"textbox exercises\">\n<h3>ExAMPLE<\/h3>\n<p>Let [latex]f(x)=3x^2-6x-2[\/latex] and [latex]g(x)=4x-1[\/latex]<\/p>\n<p>Compute the following:<\/p>\n<p>[latex]1)\\, f(-1)+g(4)\\\\[\/latex]<\/p>\n<p>[latex]2)\\, f(-1)-g(4)\\\\[\/latex]<\/p>\n<p>[latex]3)\\, g(-2)-g(3)\\\\[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q661294\">Show Solution<\/span><\/p>\n<div id=\"q661294\" class=\"hidden-answer\" style=\"display: none\">\n<p>1)<\/p>\n<p>[latex]f(-1)+g(4)[\/latex]<\/p>\n<p>To find [latex]f(-1),[\/latex] substitute [latex]\\require{color}\\color{Green}{-1}[\/latex] in place of [latex]x[\/latex] in the given function.\u00a0To find [latex]f(4)[\/latex] substitute\u00a0[latex]\\require{color}\\color{Green}{4}[\/latex] in place of [latex]x[\/latex] in the given function.<\/p>\n<p>[latex]\\begin{align}f(\\color{Green}{-1}\\color{black}{)} &=3(\\color{Green}{-1}\\color{black}{)^2-6(}\\color{Green}{-1}\\color{black}{)-2}\\\\ &= (3)(1)-6(-1)-2\\\\ &= 3+6-2\\\\ &= 7 \\end{align}[\/latex]<\/p>\n<p>and<\/p>\n<p>[latex]\\begin{align}g(\\color{Green}{4}\\color{black}{)} &=4(\\color{Green}{4}\\color{black}{)-1(}\\\\ &= 16-1\\\\ &=15 \\end{align}[\/latex]<\/p>\n<p>We now have, [latex]f(-1)=7[\/latex] and [latex]g(4)=15[\/latex].<\/p>\n<p>Let&#8217;s substitute these in place of [latex]f(-1)[\/latex] and [latex]g(4)[\/latex].<\/p>\n<p>[latex]\\require{color} f(\\color{Green}{-1}\\color{black}{)+g(}\\color{Green}{4}\\color{black}{)}[\/latex]<\/p>\n<p>[latex]\\begin{align}&= 7+15\\\\ &= 22\\end{align}\\\\\\\\[\/latex]<\/p>\n<p>2)<\/p>\n<p>[latex]f(-1)-g(4)\\\\[\/latex]<\/p>\n<p>In part 1 we found [latex]f(-1)=7[\/latex] and [latex]g(4)=15[\/latex].<\/p>\n<p>Let&#8217;s substitute these in place of [latex]f(-1)[\/latex] and [latex]g(4)[\/latex].<\/p>\n<p>[latex]f(-1)-g(4)[\/latex]<\/p>\n<p>[latex]\\begin{align}&= 7-15\\\\ &= -8\\end{align}\\\\\\\\[\/latex]<\/p>\n<p>3)<\/p>\n<p>[latex]g(-2)-g(3)\\\\[\/latex]<\/p>\n<p>To find [latex]g(-2),[\/latex] substitute [latex]\\require{color}\\color{Green}{-2}[\/latex] in place of [latex]x[\/latex] in the given function.\u00a0To find [latex]g(3)[\/latex] substitute\u00a0[latex]\\require{color}\\color{Green}{3}[\/latex] in place of [latex]x[\/latex] in the given function.<\/p>\n<p>[latex]\\begin{align}g(\\color{Green}{-2}\\color{black}{)} &=4(\\color{Green}{-2}\\color{black}{)-1}\\\\ &= -8-1\\\\ &= -9 \\end{align}[\/latex]<\/p>\n<p>and<\/p>\n<p>[latex]\\begin{align}g(\\color{Green}{3}\\color{black}{)} &=4(\\color{Green}{3}\\color{black}{)-1}\\\\ &= 12-1\\\\ &=11 \\end{align}[\/latex]<\/p>\n<p>We now have, [latex]g(-2)=-9[\/latex] and [latex]g(3)=11[\/latex].<\/p>\n<p>Let&#8217;s substitute these in place of [latex]g(-2)[\/latex] and [latex]g(3)[\/latex].<\/p>\n<p>[latex]\\require{color} g(\\color{Green}{-2}\\color{black}{)-g(}\\color{Green}{3}\\color{black}{)}[\/latex]<\/p>\n<p>[latex]\\begin{align}&= -9-11\\\\ &= -20\\end{align}\\\\\\\\[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Here is a video with more examples like the one above.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  The Sum and Difference of Two Function Values With Different Inputs\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/cqDZl1DrZIg?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Let [latex]f(x)=2x^3-5x+3[\/latex] and [latex]h(x)=x-5[\/latex],<\/p>\n<p>Find the following:<\/p>\n<p>[latex]1) (f+h)(x)[\/latex] and state the domain of the combined function.<\/p>\n<p>[latex]2) (h-f)(x)[\/latex] and state the domain of the combined function.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q295585\">Show Solution<\/span><\/p>\n<div id=\"q295585\" class=\"hidden-answer\" style=\"display: none\">\n<p>1)<\/p>\n<p>[latex]\\require{color}\\begin{align}(f+h)(x)=f(x)+ h(x) & =(2x^3-5x+3)+(x-5) \\\\ & =2x^3-5x+x+3-5 && \\color{blue}{\\textsf{group like terms}} \\\\ & =2x^3-4x-2 && \\color{blue}{\\textsf{simplify}}\\end{align}[\/latex]<\/p>\n<p>The domain of the combined function is the intersection of the domains of the original functions.<br \/>\nThe domain of [latex]f(x)=2x^3-5x+3[\/latex] is [latex](-\\infty, \\infty)[\/latex]\u00a0because there are no restrictions on the domain.<br \/>\nThe domain of [latex]h(x)=x-5[\/latex] is [latex](-\\infty, \\infty)[\/latex] because there are also no restrictions on the domain.<br \/>\nThe intersection of the two domains will also be\u00a0[latex](-\\infty, \\infty)[\/latex].<\/p>\n<p>2)<\/p>\n<p>[latex]\\require{color}\\begin{align}(h-f)(x)=h(x)-f(x) & =(x-5)-(2x^3-5x+3)\\\\ & =x-5-2x^2+5x-3 && \\color{blue}{\\textsf{distribute negative}} \\\\ & =-2x^2+6x-8 && \\color{blue}{\\textsf{simplify}}\\end{align}[\/latex]<\/p>\n<p>The domain of the combined function is the intersection of the domains of the original functions.<br \/>\nThe domain of [latex]f(x)=2x^3-5x+3[\/latex] is [latex](-\\infty, \\infty)[\/latex]\u00a0because there are no restrictions on the domain.<br \/>\nThe domain of [latex]h(x)=x-5[\/latex] is [latex](-\\infty, \\infty)[\/latex] because there are also no restrictions on the domain.<br \/>\nThe intersection of the two domains will also be\u00a0[latex](-\\infty, \\infty)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In our next example, we will evaluate a sum and difference of functions and show that you can get to the same result in one of two ways.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Let [latex]f(x)=2x^3-5x+3[\/latex] and [latex]h(x)=x-5[\/latex]<\/p>\n<p>Evaluate:\u00a0[latex](f+h)(2)[\/latex]<\/p>\n<p>Show that you get the same result by<\/p>\n<p>1) Evaluating the functions first, then performing the indicated operation on the result.<\/p>\n<p>2) Performing the operation on the functions first then evaluating the result.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q754772\">Show Solution<\/span><\/p>\n<div id=\"q754772\" class=\"hidden-answer\" style=\"display: none\">\n<p>1) First, we will evaluate the functions separately:<\/p>\n<p>[latex]\\require{color}\\begin{align}f(\\color{Green}{2}\\color{black}{)} &= 2(\\color{Green}{2}\\color{black}{)}^3-5(\\color{Green}{2}\\color{black}{)+3}\\\\ &=16-10+3\\\\ &=9 \\end{align}[\/latex]<\/p>\n<p>[latex]\\begin{align}h(\\color{Green}{2}\\color{black}{)} &= (\\color{Green}{2}\\color{black}{)-5}\\\\ &=-3 \\end{align}[\/latex]<\/p>\n<p>Now we will perform the indicated operation using the results:<\/p>\n<p>[latex]\\begin{align}(f+h)(2) &= f(2)+h(2)\\\\ &= 9+(-3)\\\\ &=6 \\end{align}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>2) We can get the same result by adding the functions first and then evaluating the result at\u00a0[latex]x=2[\/latex].<\/p>\n<p>[latex]\\begin{align} (f+h)(x) &=f(x)+h(x)\\\\ &=(2{x}^{3}-5x+3)+(x-5)\\\\ &=2x^3-4x-2 \\end{align}[\/latex].<\/p>\n<p>Now we can evaluate this result at\u00a0[latex]x=2[\/latex]<\/p>\n<p>[latex]\\require{color}\\begin{align} (f+h)(\\color{Green}{2}\\color{black}{)} &=2(\\color{Green}{2}\\color{black}{)^3-4(}\\color{Green}{2}\\color{black}{)-2}\\\\ &=16-8-2\\\\ &=6 \\end{align}[\/latex]<\/p>\n<p>Both methods give the same result, and both require about the same amount of work.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>ExAMPLE<\/h3>\n<p>Given: [latex]f(x)=x^2-3x\\\\[\/latex]<\/p>\n<p>Find:<\/p>\n<p>[latex]1)f(x+1)-f(2)\\\\[\/latex]<\/p>\n<p>[latex]2)f(x+1)+f(x)\\\\[\/latex]<\/p>\n<p>[latex]3)f(x+h)-f(x)\\\\[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q830895\">Show Solution<\/span><\/p>\n<div id=\"q830895\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]1)f(x+1)-f(2)\\\\[\/latex]<\/p>\n<p>To find [latex]f(x+1),[\/latex] substitute [latex]\\require{color}\\color{Green}{x+1}[\/latex] in place of [latex]x[\/latex] in the given function.\u00a0To find [latex]f(2)[\/latex] substitute\u00a0[latex]\\require{color}\\color{Green}{2}[\/latex] in place of [latex]x[\/latex] in the given function.<\/p>\n<p>[latex]\\require{color}\\begin{align}f(x) &= x^2-3x\\\\f(\\color{Green}{x+1}\\color{black}{)}&= (\\color{Green}{x+1}\\color{black}{)^{2}-3}(\\color{Green}{x+1}\\color{black}{)}\\\\ f(\\color{Green}{2}\\color{black}{)} &=(\\color{Green}{2}\\color{black}{)}^{2}\\color{black}{-3(}\\color{Green}{2}\\color{black}{)}\\end{align}[\/latex]<\/p>\n<p>We now have, [latex]f(x+1)=\\color{BurntOrange}{{(x+1)}^{2}-3(x+1)}[\/latex] and [latex]f(2)=\\color{Plum}{(2)^2-3(2)}[\/latex].<\/p>\n<p>Let&#8217;s substitute these in place of [latex]f(x+1)[\/latex] and [latex]f(2)[\/latex].<\/p>\n<p>[latex]\\require{color}\\color{BurntOrange}{f(x+1)}\\color{black}{+}\\color{Plum}{f(2)}[\/latex]<\/p>\n<p>[latex]\\require{color}\\begin{align} &=\\color{BurntOrange}{{(x+1)}^{2}-3(x+1)}\\color{black}{+}\\color{Plum}{(2)^2-3(2)} &&\\color{blue}{\\textsf{substitute}}\\\\ &= \\color{BurntOrange}{(x+1)(x+1)-3(x+1)}\\color{black}{+}\\color{Plum}{(2)(2)-3(2)}\\\\ &= \\color{BurntOrange}{x^2+2x+1-3x-3}\\,\\color{black}{+} \\color{Plum}{\\,4-6} &&\\color{blue}{\\textsf{multiply}}\\\\ &= \\color{BurntOrange}{x^2-x-2}\\,\\color{black}{+}\\,\\color{Plum}{(-2)} &&\\color{blue}{\\textsf{combine like terms}}\\\\ &= x^2-x-4 &&\\color{blue}{\\textsf{combine like terms}}\\end{align}\\\\[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>[latex]2)f(x+1)+f(x)\\\\[\/latex]<\/p>\n<p>To find [latex]f(x+1),[\/latex] substitute [latex]\\require{color}\\color{Green}{x+1}[\/latex] in place of [latex]x[\/latex] in the given function.<\/p>\n<p>[latex]\\require{color}\\begin{align}f(x) &= x^2-3x\\\\f(\\color{Green}{x+1}\\color{black}{)} &= (\\color{Green}{x+1}\\color{black}{)^{2}-3}(\\color{Green}{x+1}\\color{black}{)}\\end{align}[\/latex]<\/p>\n<p>We now have, [latex]f(x+1)=\\color{BurntOrange}{{(x+1)}^{2}-3(x+1)}[\/latex] and [latex]f(x)=\\color{Plum}{x^2-3x}[\/latex].<\/p>\n<p>Let&#8217;s substitute these in place of [latex]f(x+1)[\/latex] and [latex]f(x)[\/latex].<\/p>\n<p>[latex]\\require{color}\\color{BurntOrange}{f(x+1)}\\color{black}{+}\\color{Plum}{f(x)}[\/latex]<\/p>\n<p>[latex]\\require{color}\\begin{align} &=\\color{BurntOrange}{{(x+1)}^{2}-3(x+1)}\\color{black}{+}\\color{Plum}{x^2-3x} &&\\color{blue}{\\textsf{substitute}}\\\\ &= \\color{BurntOrange}{(x+1)(x+1)-3(x+1)}\\color{black}{+}\\color{Plum}{x^2-3x}\\\\ &= \\color{BurntOrange}{x^2+2x+1-3x-3}\\color{black}{+}\\color{Plum}{x^2-3x} &&\\color{blue}{\\textsf{multiply}}\\\\ &= 2x^2-4x-2 &&\\color{blue}{\\textsf{combine like terms}}\\end{align}\\\\[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>[latex]3)f(x+h)-f(x)\\\\[\/latex]<\/p>\n<p>To find [latex]f(x+h),[\/latex] substitute [latex]\\require{color}\\color{Green}{x+h}[\/latex] in place of [latex]x[\/latex] in the given function.<\/p>\n<p>[latex]\\require{color}\\begin{align}f(x) &= x^2-3x\\\\f(\\color{Green}{x+h}\\color{black}{)} &= (\\color{Green}{x+h}\\color{black}{)^{2}-3}(\\color{Green}{x+h}\\color{black}{)}\\end{align}[\/latex]<\/p>\n<p>We now have, [latex]f(x+h)=\\color{BurntOrange}{{(x+h)}^{2}-3}(x+h)[\/latex] and [latex]f(x)=\\color{Plum}{x^2-3x}[\/latex].<\/p>\n<p>Let&#8217;s substitute these in place of [latex]f(x+h)[\/latex] and [latex]f(x)[\/latex].<\/p>\n<p>[latex]\\require{color}\\color{BurntOrange}{f(x+h)}\\color{black}{-}\\color{Plum}{f(x)}[\/latex]<\/p>\n<p>[latex]\\require{color}\\begin{align} &=\\color{BurntOrange}{{(x+h)}^{2}-3(x+h)}\\color{black}{-}\\color{Plum}{(x^2-3x)} &&\\color{blue}{\\textsf{substitute}}\\\\ &= \\color{BurntOrange}{(x+h)(x+h)-3(x+h)}\\color{black}{-}\\color{Plum}{(x^2-3x)}\\\\ &= \\color{BurntOrange}{x^2+2xh+h^2-3x-3h}\\color{Plum}{-x^2+3x} &&\\color{blue}{\\textsf{distributive property}}\\\\ &= 2xh+h^2-3h &&\\color{blue}{\\textsf{combine like terms}}\\end{align}\\\\[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Multiply and Divide Polynomial Functions<\/h2>\n<p>To multiply and divide polynomial functions, we can either evaluate the function first and then perform the indicated operation or vice-versa.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Let [latex]g(t)=t+7[\/latex] and [latex]f(t)=5t^2-3[\/latex]<\/p>\n<p>1) Find [latex](g \u00b7 f)(t)[\/latex]<\/p>\n<p>2) Evaluate [latex](g \u00b7 f)(-1)[\/latex]<\/p>\n<p>3) Evaluate [latex](f \\cdot g)(-1)[\/latex]<\/p>\n<p>Should Part 2 and Part 3 have the same answer? Explain.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q48983\">Show Solution<\/span><\/p>\n<div id=\"q48983\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">1)<\/p>\n<p style=\"text-align: left;\">[latex]\\begin{align}\\require{color}(g \u00b7 f)(t) & =(t+7)(5t^2-3) \\\\ & =t\\cdot(5t^2)+t\\cdot(-3)+7\\cdot(5t^2)+7\\cdot(-3) &&\\color{blue}{\\textsf{apply the distributive property}} \\\\ & =5t^3-3t+35t^2-21 &&\\color{blue}{\\textsf{simplify}}\\\\ & =5t^3+35t^2-3t-21 &&\\color{blue}{\\textsf{write in descending order}}\\end{align}\\\\[\/latex]<\/p>\n<p>2)<\/p>\n<p>Evaluate\u00a0[latex](g \u00b7 f)(-1)[\/latex]<\/p>\n<p style=\"text-align: left;\">[latex]\\begin{align}\\require{color}(g \u00b7 f)(t) &=5t^3+35t^2-3t-21 && \\color{blue}{\\textsf{from above}}\\\\(g \u00b7 f)(-1) &=5(-1)^3+35(-1)^2-3(-1)-21 && \\color{blue}{\\textsf{evaluate exponents and multiply}}\\\\ &=-5+35+3-21 && \\color{blue}{\\textsf{add and subtract}}\\\\ &=12\\end{align}[\/latex]<\/p>\n<p>3)<\/p>\n<p>Evaluate [latex](f \\cdot g)(-1)[\/latex]<\/p>\n<p>We could take the same steps as we did above by finding\u00a0[latex](f \\cdot g)(t)[\/latex] and then substituting [latex]-1[\/latex] in place of [latex]t[\/latex]. Here is another way you can evaluate\u00a0[latex](f \\cdot g)(-1)[\/latex].<\/p>\n<p>[latex](f \\cdot g)(-1)= f(-1) \\cdot g(-1)[\/latex]<\/p>\n<p>Let&#8217;s first find [latex]f(-1)[\/latex] and [latex]g(-1)[\/latex]. Then multiply the results together.<\/p>\n<p>[latex]\\begin{align}f(\\color{Green}{-1}\\color{black}{)} &= 5(\\color{Green}{-1}\\color{black}{)^2-3}\\\\ &= 5(1)-3\\\\ &= 5-3\\\\ &= 2 \\end{align}[\/latex]<\/p>\n<p>[latex]\\begin{align}g(\\color{Green}{-1}\\color{black}{)} &= \\color{Green}{-1}\\color{black}{+7}\\\\ &= 6\\end{align}[\/latex]<\/p>\n<p>Multiply results together.<\/p>\n<p>[latex](f \\cdot g)(-1) = f(-1) \\cdot g(-1) = (2)\\cdot (6) = 12[\/latex]<\/p>\n<p>Part 2 and Part 3 have the same answer because multiplication is commutative.<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next example, we will divide polynomial functions and then evaluate the new function.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Given [latex]p(x)=4x[\/latex] and [latex]r(x)=x+3[\/latex]<\/p>\n<p>Find [latex]\\left(\\dfrac{p}{r}\\right)(x)[\/latex] and state its domain. Then evaluate\u00a0[latex]\\left(\\dfrac{p}{r}\\right)(-2)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q374371\">Show Solution<\/span><\/p>\n<div id=\"q374371\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(\\dfrac{p}{r}\\right)(x)=\\dfrac{p(x)}{r(x)}=\\dfrac{4x}{x+3},\\\\[\/latex]<\/p>\n<p>Domain: All real numbers except [latex]x\\not=-3.[\/latex]The value [latex]x=-3[\/latex] will make the denominator equal to zero and therefore the expression will be undefined.[latex]\\\\[\/latex]<\/p>\n<p>Domain in Interval Notation: [latex](-\\infty, -3)\\bigcup(-3, \\infty)\\\\[\/latex]<\/p>\n<p>Domain in Set-Builder Notation: [latex]\\{ x | x\\not=-3 \\}\\\\\\\\[\/latex]<\/p>\n<p>Evaluate this quotient for\u00a0[latex]x = -2[\/latex].<\/p>\n<p>[latex]\\begin{align}\\left(\\dfrac{p}{r}\\right)(-2) &=\\dfrac{p(-2)}{r(-2)}\\\\\\\\ &=\\dfrac{4(-2)}{-2+3}\\\\\\\\ &=\\dfrac{-8}{1}\\\\\\\\ &=-8\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In our next example we will find function values from a graph.<\/p>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE<\/h3>\n<p>For the functions [latex]\\color{DarkRed}{f(x)}[\/latex], red graph, and [latex]\\color{blue}{g(x)}[\/latex], blue graph, pictured, find the function values.<\/p>\n<p>[latex]1)f(-1)[\/latex]<\/p>\n<p>[latex]2)(f-g)(2)[\/latex]<\/p>\n<p>[latex]3)(f+g)(0)[\/latex]<\/p>\n<p>[latex]4)\\left(\\dfrac{f}{g}\\right)(2)\\\\[\/latex]<\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/read-algebra-of-polynomial-functions\/desmos-graph-31\/\" rel=\"attachment wp-att-1574\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1574\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/desmos-graph-31.png\" alt=\"Two curves are shown: red curve increasing going through (negative 1, negative 2), (0, negative 1), (1,0) and (2,7). Blue curve increasing through (0, negative 5), (1, negative 2), (2,1) and (9,4).\" width=\"454\" height=\"454\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/desmos-graph-31.png 800w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/desmos-graph-31-150x150.png 150w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/desmos-graph-31-300x300.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/desmos-graph-31-768x768.png 768w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/desmos-graph-31-65x65.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/desmos-graph-31-225x225.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/desmos-graph-31-350x350.png 350w\" sizes=\"auto, (max-width: 454px) 100vw, 454px\" \/><\/a><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q597368\">Show Solution<\/span><\/p>\n<div id=\"q597368\" class=\"hidden-answer\" style=\"display: none\">\n<p>1)[latex]\\begin{align}\\require{color}f(-1)=-2 &&\\color{blue}{\\textsf{when the input is -1, the output is -2}} \\end{align}\\\\[\/latex]<\/p>\n<p>2)[latex]\\begin{align}\\require{color}\\left(f-g \\right)(2) &=f(2)-g(2)\\\\ &=7-1 &&\\color{blue}{\\textsf{subtract}}\\\\ &=6 \\end{align}\\\\[\/latex]<\/p>\n<p>3)[latex]\\begin{align}\\require{color}\\left(f+g \\right)(0) &=f(0)+g(0)\\\\ &=-1+(-5) &&\\color{blue}{\\textsf{add}}\\\\ &=-6 \\end{align}\\\\[\/latex]<\/p>\n<p>4)[latex]\\left(\\dfrac{f}{g}\\right)(2)=\\dfrac{f(2)}{g(2)}=\\dfrac{7}{1}=7[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Now try this next example:<\/p>\n<div class=\"textbox tryit\"><iframe loading=\"lazy\" id=\"ohm288649\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288649&theme=oea&iframe_resize_id=ohm288649&sameseed=1&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/div>\n","protected":false},"author":395986,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"d451c5c9-43d1-4a9a-ad5b-868823952373","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-41","chapter","type-chapter","status-publish","hentry"],"part":143,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/41","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/users\/395986"}],"version-history":[{"count":51,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/41\/revisions"}],"predecessor-version":[{"id":2115,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/41\/revisions\/2115"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/parts\/143"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/41\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/media?parent=41"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=41"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/contributor?post=41"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/license?post=41"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}