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}\begin{align}{5n}&\hspace{2mm}\text{if}\hspace{2mm}{0}<{n}<{10}\\ 50&\hspace{2mm}\text{if}\hspace{2mm}{n}\ge 10\end{align}\end{cases}[/latex]

Analysis of the Solution

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.

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.

[latex]C(4)=25 + 10( 4-2) =$45[/latex]

Analysis of the Solution

We can see where the function changes from a constant to a shifted and stretched identity at [latex]g=2[/latex]. We plot the graphs for the different formulas on a common set of axes, making sure each formula is applied on its proper domain.

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.