Question

1. given the graph of y = f(x), sketch the graph of its derivative y = f(x).

Explanation:

Step1: Recall derivative - slope relationship

The derivative $y = f'(x)$ at a point is the slope of the tangent line to the curve $y = f(x)$ at that point.

Step2: Analyze intervals of increase and decrease for (a)

When $y = f(x)$ is increasing, $f'(x)>0$. When it is decreasing, $f'(x)<0$. At local - maxima and minima, $f'(x) = 0$. For the graph in (a), it is decreasing on $(-3,-2)$ and $(0,2)$, so $f'(x)<0$ on these intervals. It is increasing on $(-2,0)$ and $(2,3)$, so $f'(x)>0$ on these intervals. There are local minima at $x=-2$ and $x = 2$ and a local maximum at $x = 0$, so $f'(x)=0$ at $x=-2,0,2$.

Step3: Analyze intervals of increase and decrease for (b)

For the graph in (b), it is increasing on $(-3,-1)$ and $(1,3)$, so $f'(x)>0$ on these intervals. It is decreasing on $(-1,1)$, so $f'(x)<0$ on this interval. There is a local maximum at $x=-1$ and a local minimum at $x = 1$, so $f'(x)=0$ at $x=-1,1$.

Step4: Analyze intervals of increase and decrease for (c)

For the graph in (c), it is increasing on $(-4,-2)$, so $f'(x)>0$ on $(-4,-2)$. It is constant on $(-2,2)$, so $f'(x)=0$ on $(-2,2)$. It is decreasing on $(2,4)$, so $f'(x)<0$ on $(2,4)$.

Answer:

Sketch the derivative graphs for each part as follows:
For (a): $f'(x)$ is negative on $(-3,-2)$ and $(0,2)$, positive on $(-2,0)$ and $(2,3)$, and zero at $x=-2,0,2$.
For (b): $f'(x)$ is negative on $(-1,1)$, positive on $(-3,-1)$ and $(1,3)$, and zero at $x=-1,1$.
For (c): $f'(x)$ is positive on $(-4,-2)$, zero on $(-2,2)$, and negative on $(2,4)$. (Note: A full - scale sketch would require plotting points based on estimated slopes at various $x$ - values within the intervals, but the above describes the general shape and sign changes of $f'(x)$ for each part.)