Answer
a. $x=\pm2$
b. $x=\pm1$
c. $x=0$
d. Even,
e. No,
f. $f(0)=-2$, relative minimum
g. See graph (red curve)
h. See graph (blue curve)
i. See graph (green curve)
j. $-\frac{1}{3}$
Work Step by Step
Use the graph given in the exercise, we have:
a. The zeros of $f$ can be identified as $x=\pm2$
b. We can identify the x-values as $x=\pm1$ such that $f(x)=-1$
c. We can identify the x-values as $x=0$ such that $f(x)=-2$
d. Even, because the function is symmetric around the y-axis.
e. No, because it will fail the horizontal line test.
f. $f(0)=-2$ is a relative minimum because we can find an open interval around $x=0$ such that $f(x)\gt f(0)$ when $x\ne0$.
g. See graph (red curve)
h. See graph (blue curve)
i. See graph (green curve)
j. We have $f(-2)=0$ and $f(1)=-1$; thus the rate of change is $\frac{-1-0}{1+2}=-\frac{1}{3}$
