CPM Homework Help : CALC3RD Problem 8-86

Multiple Choice: The graph of $y = g\left(x\right)$ shown below has horizontal tangents at $\left(–1, 1\right)$ and $\left(1, –1\right)$ . It has vertical tangent at $\left(0, 0\right)$ . For what values of $x$ on the open interval $\left(–2, 4\right)$ is $g$ not differentiable?

$2$ only

$0$ and $2$

$−1$ and $1$

$0$ only

$−1, 0, 1$ , and $2$

Piecewise labeled, g of x, left semicircle with vertices at (negative 2, comma 0), (negative 1, comma 1), & the origin, center semicircle, vertices at the origin, (1, comma negative 1), & (2, comma 0), left increasing concave up curve, starting at open point (2, comma negative 2), passing through the point (4, comma 4).

Hint:

First of all, differentiability implies continuity. (That is, a point of discontinuity does not have a derivative.) Secondly, the slopes must agree from the left and the right.

$\left (\text{That is to say, }\lim \limits_{x\rightarrow a^{-}}g'(x)=\lim \limits_{x\rightarrow a^{+}}g'(x) \right )$

Next, the value of the derivative must exist. (That is, the derivative must be finite.) Lastly, the value of the derivative must agree with the limit of the derivative at that point.