1)

[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]

2)

[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]

1)

[latex] f(-1)+g(4)[/latex]

To find [latex] f(-1), [/latex] substitute [latex] \require{color}\color{Green}{-1} [/latex] in place of [latex] x [/latex] in the given function. To find [latex] f(4) [/latex] substitute [latex] \require{color}\color{Green}{4} [/latex] in place of [latex] x [/latex] in the given function.

[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]

and

[latex] \begin{align}g(\color{Green}{4}\color{black}{)} &=4(\color{Green}{4}\color{black}{)-1(}\\ &= 16-1\\ &=15 \end{align} [/latex]

We now have, [latex] f(-1)=7 [/latex] and [latex] g(4)=15 [/latex].

Let’s substitute these in place of [latex] f(-1) [/latex] and [latex] g(4) [/latex].

[latex]\require{color} f(\color{Green}{-1}\color{black}{)+g(}\color{Green}{4}\color{black}{)}[/latex]

[latex]\begin{align}&= 7+15\\ &= 22\end{align}\\\\[/latex]

2)

[latex] f(-1)-g(4)\\ [/latex]

In part 1 we found [latex] f(-1)=7 [/latex] and [latex] g(4)=15 [/latex].

Let’s substitute these in place of [latex] f(-1) [/latex] and [latex] g(4) [/latex].

[latex] f(-1)-g(4)[/latex]

[latex]\begin{align}&= 7-15\\ &= -8\end{align}\\\\[/latex]

3)

[latex] g(-2)-g(3)\\ [/latex]

To find [latex] g(-2), [/latex] substitute [latex] \require{color}\color{Green}{-2} [/latex] in place of [latex] x [/latex] in the given function. To find [latex] g(3) [/latex] substitute [latex] \require{color}\color{Green}{3} [/latex] in place of [latex] x [/latex] in the given function.

[latex] \begin{align}g(\color{Green}{-2}\color{black}{)} &=4(\color{Green}{-2}\color{black}{)-1}\\ &= -8-1\\ &= -9 \end{align} [/latex]

and

[latex] \begin{align}g(\color{Green}{3}\color{black}{)} &=4(\color{Green}{3}\color{black}{)-1}\\ &= 12-1\\ &=11 \end{align} [/latex]

We now have, [latex] g(-2)=-9 [/latex] and [latex] g(3)=11 [/latex].

Let’s substitute these in place of [latex] g(-2) [/latex] and [latex] g(3) [/latex].

[latex]\require{color} g(\color{Green}{-2}\color{black}{)-g(}\color{Green}{3}\color{black}{)}[/latex]

[latex]\begin{align}&= -9-11\\ &= -20\end{align}\\\\[/latex]

1)

[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]

The domain of the combined function is the intersection of the domains of the original functions.
The domain of [latex]f(x)=2x^3-5x+3[/latex] is [latex] (-\infty, \infty) [/latex] because there are no restrictions on the domain.
The domain of [latex] h(x)=x-5 [/latex] is [latex] (-\infty, \infty) [/latex] because there are also no restrictions on the domain.
The intersection of the two domains will also be [latex] (-\infty, \infty) [/latex].

2)

[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]

The domain of the combined function is the intersection of the domains of the original functions.
The domain of [latex]f(x)=2x^3-5x+3[/latex] is [latex] (-\infty, \infty) [/latex] because there are no restrictions on the domain.
The domain of [latex] h(x)=x-5 [/latex] is [latex] (-\infty, \infty) [/latex] because there are also no restrictions on the domain.
The intersection of the two domains will also be [latex] (-\infty, \infty) [/latex].

1) First, we will evaluate the functions separately:

[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]

[latex]\begin{align}h(\color{Green}{2}\color{black}{)} &= (\color{Green}{2}\color{black}{)-5}\\ &=-3 \end{align}[/latex]

Now we will perform the indicated operation using the results:

[latex]\begin{align}(f+h)(2) &= f(2)+h(2)\\ &= 9+(-3)\\ &=6 \end{align}[/latex]

2) We can get the same result by adding the functions first and then evaluating the result at [latex]x=2[/latex].

[latex]\begin{align} (f+h)(x) &=f(x)+h(x)\\ &=(2{x}^{3}-5x+3)+(x-5)\\ &=2x^3-4x-2 \end{align}[/latex].

Now we can evaluate this result at [latex]x=2[/latex]

[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]

Both methods give the same result, and both require about the same amount of work.

[latex]1)f(x+1)-f(2)\\[/latex]

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. To find [latex] f(2) [/latex] substitute [latex] \require{color}\color{Green}{2} [/latex] in place of [latex] x [/latex] in the given function.

[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]

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].

Let’s substitute these in place of [latex] f(x+1) [/latex] and [latex] f(2) [/latex].

[latex] \require{color}\color{BurntOrange}{f(x+1)}\color{black}{+}\color{Plum}{f(2)} [/latex]

[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]

[latex]2)f(x+1)+f(x)\\[/latex]

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.

[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]

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].

Let’s substitute these in place of [latex] f(x+1) [/latex] and [latex] f(x) [/latex].

[latex] \require{color}\color{BurntOrange}{f(x+1)}\color{black}{+}\color{Plum}{f(x)} [/latex]

[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]

[latex]3)f(x+h)-f(x)\\[/latex]

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.

[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]

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].

Let’s substitute these in place of [latex] f(x+h) [/latex] and [latex] f(x) [/latex].

[latex] \require{color}\color{BurntOrange}{f(x+h)}\color{black}{-}\color{Plum}{f(x)} [/latex]

[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]

1)

[latex]\begin{align}\require{color}(g · 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]

2)

Evaluate [latex](g · f)(-1)[/latex]

[latex]\begin{align}\require{color}(g · f)(t) &=5t^3+35t^2-3t-21 && \color{blue}{\textsf{from above}}\\(g · 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]

3)

Evaluate [latex] (f \cdot g)(-1) [/latex]

We could take the same steps as we did above by finding [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 [latex] (f \cdot g)(-1) [/latex].

[latex] (f \cdot g)(-1)= f(-1) \cdot g(-1) [/latex]

Let’s first find [latex] f(-1) [/latex] and [latex] g(-1) [/latex]. Then multiply the results together.

[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]

[latex] \begin{align}g(\color{Green}{-1}\color{black}{)} &= \color{Green}{-1}\color{black}{+7}\\ &= 6\end{align} [/latex]

Multiply results together.

[latex] (f \cdot g)(-1) = f(-1) \cdot g(-1) = (2)\cdot (6) = 12 [/latex]

Part 2 and Part 3 have the same answer because multiplication is commutative.

[latex]\left(\dfrac{p}{r}\right)(x)=\dfrac{p(x)}{r(x)}=\dfrac{4x}{x+3},\\[/latex]

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]

Domain in Interval Notation: [latex] (-\infty, -3)\bigcup(-3, \infty)\\ [/latex]

Domain in Set-Builder Notation: [latex] \{ x | x\not=-3 \}\\\\ [/latex]

Evaluate this quotient for [latex]x = -2[/latex].

[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]

1)[latex]\begin{align}\require{color}f(-1)=-2 &&\color{blue}{\textsf{when the input is -1, the output is -2}} \end{align}\\[/latex]

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]

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]

4)[latex]\left(\dfrac{f}{g}\right)(2)=\dfrac{f(2)}{g(2)}=\dfrac{7}{1}=7[/latex]