Chapter 3.1, Problem 22E

To determine

Tograph:The one graph which is the derivative of f for $x>0$ .

Answer to Problem 22E

Explanation of Solution

Given information:

The graph:

  

Find the points on the graph of the derivative $f\text{'}\left(x\right)$ .

The derivative $f\text{'}\left(x\right)$ will have x intercepts where the slop of $f\left(x\right)$ is zero.

From the graph it can be seen that at $x=-2$ and $x=1$ the slop of $f\left(x\right)$ is zero.

Thus, the derivative $f\text{'}\left(x\right)$ has the value $y=0$ at $x=-2$ and $x=1$ .

The derivative of $f\text{'}\left(x\right)$ will be positive where the curve of $f\left(x\right)$ is increases. Similarly the derivative of $f\text{'}\left(x\right)$ will be negative where the curve of $f\left(x\right)$ is decreases.

Thus from the given graph, from the interval $\left(-\infty ,-2\right)$ to $\left(1,\infty \right)$ the derivative of $f\text{'}\left(x\right)$ will be positive. And the derivative will be negative on $\left(-2,1\right)$ .

The derivative will have a vertex where the slop of $f\left(x\right)$ is at steepest negative value.

At point $x=-\frac{1}{2}$ , the slop of $f\left(x\right)$ has the steepest negative value.

Consider the points $\left(-1,1\right)$ and $\left(0,-1\right)$ .

Now, calculate the slop of the slop $\left(-1,1\right)$ and $\left(0,-1\right)$ , use the slop formula of straight line.

  $\begin{array}{c}{m}_{\min }=\frac{-1-1}{0-\left(-1\right)}\\ =-2\end{array}$

Thus, $f\text{'}\left(-\frac{1}{2}\right)=-2$

Draw the graph of the derivative.

From the above information draw the graph of the derivative: