Question

1. (3 points) the graph of h is shown below. sketch the graph of h.

Explanation:

Step1: Recall derivative - slope relationship

The derivative $h'(x)$ represents the slope of the tangent line to the graph of $y = h(x)$ at a point $x$.

Step2: Analyze intervals of increasing and decreasing

When $h(x)$ is increasing, $h'(x)>0$. When $h(x)$ is decreasing, $h'(x)<0$. From the graph of $h(x)$, it is increasing on $(-2,- 1)$ and $(3,5)$, so $h'(x)>0$ on these intervals. It is decreasing on $(-1,3)$, so $h'(x)<0$ on this interval.

Step3: Identify critical points

The critical points of $h(x)$ are at $x=-1$ and $x = 3$ (where the slope of the tangent is 0). So $h'(-1)=0$ and $h'(3)=0$.

Step4: Sketch the graph of $h'$

Start by plotting the points $(-1,0)$ and $(3,0)$. Draw a positive - valued curve for $x\in(-2,-1)\cup(3,5)$ and a negative - valued curve for $x\in(-1,3)$.

Answer:

A sketch of the graph of $h'$ with positive values on $(-2,-1)\cup(3,5)$, negative values on $(-1,3)$ and $h'(-1) = 0$, $h'(3)=0$.