1.[latex]f(x)[/latex] is defined as [latex]7x+3[/latex] for [latex]x=-1\text{ becuase }-1<0[/latex].  

Evaluate: [latex]f(-1)=7(-1)+3=-7+3=-4[/latex]

2. [latex]f(x)[/latex] is defined as [latex]7x+6[/latex] for [latex]x=0\text{ becuase }0\ge{0}[/latex].

Evaluate: [latex]f(0)=7(0)+6=0+6=6[/latex]

3. [latex]f(x)[/latex] is defined as [latex]7x+6[/latex] for [latex]x=2\text{ becuase }2\ge{0}[/latex].

Evaluate: [latex]f(2)=7(2)+6=14+6=20[/latex]

To find the cost of using 1.5 gigabytes of data, [latex]C(1.5)[/latex], we first look to see which part of the domain our input falls in. Because 1.5 is less than 2, we use the first formula.

[latex]C(1.5) = \$25[/latex]

To find the cost of using 4 gigabytes of data, [latex]C(4)[/latex], we see that our input of 4 is greater than 2, so we use the second formula.

Two different formulas will be needed. For [latex]n[/latex]-values under 10, [latex]C=5n[/latex]. For values of [latex]n[/latex] that are 10 or greater, [latex]C=50[/latex].

[latex]C(n)=\begin{cases}{5n}\text{ if }{0}<{n}<{10}\\ 50\text{ if }{n}\ge 10\end{cases}[/latex]

Analysis of the Solution

The function is represented in the figure below. The graph is a diagonal line from [latex]n=0[/latex] to [latex]n=10[/latex] and a constant after that. In this example, the two formulas agree at the meeting point where [latex]n=10[/latex], but not all piecewise functions have this property.

First, graph the line [latex]f(x) = −x−3[/latex], erasing the part where [latex]x[/latex] is greater than [latex]-3[/latex]. Place an open circle at [latex](-3,0)[/latex].

Now place the line [latex]f(x) = x+3[/latex] on the graph, starting at the point [latex](-3,0)[/latex]. Note that for this portion of the graph, the point [latex](-3,0)[/latex] is included, so you can remove the open circle.

The two graphs meet at the point [latex](-3,0)[/latex]

The domain of this function is all real numbers because [latex](-3,0)[/latex] is not included as the endpoint of [latex]f(x) = −x−3[/latex], but it is included as the endpoint for [latex]f(x) = x+3[/latex].

The range of this function starts at [latex]f(x)=0[/latex] and includes 0,  and goes to infinity, so we would write this as [latex]x\ge0[/latex]

First, draw the line [latex]S(n)=1.5n+2.5[/latex].  We can use transformations: this is a vertical stretch of the identity by a factor of 1.5, and a vertical shift by 2.5.

S(n)=1.5n+2.5

Now we can eliminate the portions of the graph that are not in the domain based on [latex]1\le{n}\le14[/latex]

S(n) = 1.5n+2.5 for 1<=n<=14

Last, add the constant function [latex]S(n)=0[/latex] for inputs greater than or equal to 15. Place closed dots on the ends of the graph to indicate the inclusion of the end points.

Each of the component functions is from our library of toolkit functions, so we know their shapes. We can imagine graphing each function and then limiting the graph to the indicated domain. At the endpoints of the domain, we draw open circles to indicate where the endpoint is not included because of a less-than or greater-than inequality; we draw a closed circle where the endpoint is included because of a less-than-or-equal-to or greater-than-or-equal-to inequality.

Below are the three components of the piecewise function graphed on separate coordinate systems.

(a) [latex]f\left(x\right)={x}^{2}\text{ if }x\le 1[/latex]; (b) [latex]f\left(x\right)=3\text{ if 1< }x\le 2[/latex]; (c) [latex]f\left(x\right)=x\text{ if }x>2[/latex]

Now that we have sketched each piece individually, we combine them in the same coordinate plane.

Analysis of the Solution

Note that the graph does pass the vertical line test even at [latex]x=1[/latex] and [latex]x=2[/latex] because the points [latex]\left(1,3\right)[/latex] and [latex]\left(2,2\right)[/latex] are not part of the graph of the function, though [latex]\left(1,1\right)[/latex] and [latex]\left(2,3\right)[/latex] are.