Single Variable Calculus, Chapter 1, 1.1, Section 1.1, Problem 49

The graph of the given curve is shown below. Find an expression for its function.










You can find the equation of the line using the point slope form.



$\begin{equation} \begin{aligned} y - 3 &= \frac{0-3}{3-0}(x-0) && (\text{Simplifying the equation})\\ \\ y - 3 &= - x && (\text{Solving for } y)\\ \\ y &= -x + 3 \text{ for } 0 \leq x \leq 3\\ \end{aligned} \end{equation}$




Again, using point slope form, we can get the equation of the other line.



$\begin{equation} \begin{aligned} y - 0 &= \frac{4 - 0}{5 - 3}(x - 3) && (\text{Simplifying the equation and } \text{solving for } y) \\ \\ y &= 2x - 6 \text{ for } 3 < x \leq 5 && \\ \end{aligned} \end{equation}$



Therefore the final expression for this function is...



$\begin{equation} \begin{aligned} \fbox{$ f(x) = \begin{array}{ccc} -x + 3 & \text { for } & 0 \leq x \leq 3 \\ 2x - 6 & \text { for } & 3 \leq x \leq 5 \end{array} $} \end{aligned} \end{equation}$