Chapter 3.1, Problem 28E

To determine

To sketch:The graph of a continuous function $f$ with $f\left(0\right)=1$ .

Explanation of Solution

Given information:

The function: $f\text{'}\left(x\right)=\{\begin{array}{l}2,x<2\\ -1,x>2\end{array}$

Value of $x=0$

Formula Used:

Point slop form of a straight line:

  $y-y_1=m\left(x-x_1\right)$

Here $m$ is the slop of the line and $\left(x_1,y_1\right)$ is a point on the line.

Calculation:

For $x<2$ the function is $f\text{'}\left(x\right)=2$ .

Since $\frac{d}{dx}f\left(x\right)=f\text{'}\left(x\right)$ , represent the slop of a straight line.

Thus, find the expression of the straight line for the point $\left(0,1\right)$ with slop 2. Use the point slop formula $y-y_1=m\left(x-x_1\right)$

  $\begin{array}{c}y-1=2\left(x-0\right)\\ y-1=2x\\ y=2x+1\end{array}$  ......(i)

Now, find the value of $y$ , substitute the value $2$ for $x$ in equation (i).

  $\begin{array}{c}y=2\times 2+1\\ =4+1\\ =5\end{array}$

Thus, for $x>2$ , the slop of the function is $-1$ passes through the point $\left(2,5\right)$ .

Now, find the equation of the straight line. Use the point slop formula $y-y_1=m\left(x-x_1\right)$ .

  $\begin{array}{c}y-5=-1\times \left(x-2\right)\\ y-5=-x+2\\ y=x+7\end{array}$

Thus, the function is $f\left(x\right)=\{\begin{array}{l}2x+1,\text{if}x<2\\ -x+7,\text{if}x\ge 2\end{array}$

Now, draw the graph