### Home > APCALC > Chapter 5 > Lesson 5.2.5 > Problem5-103

5-103.

FUNKY FUNCTIONS REVISITED 5-103 HW eTool (Desmos). Homework Help ✎

1. Graph $f(x)=|x^3+0.125|$ and rewrite $f$ as a piecewise-defined function.

There will be two pieces. The graph indicates that $x = −0.5$ is the boundary point.

2. Zoom in at $x = –0.5$ on your graphing calculator and carefully examine the curve. Does $f$ appear differentiable at $x = –0.5$? Why or why not?

Does the slope from the left of $x = −0.5$ appear to agree with the slope from the right?

3. To confirm whether or not $f(x)=|x^3+0.125|$ is differentiable at $x = –0.5$, examine $f^\prime$. Use the piecewise-defined function from part (a) to write an equation for $f^\prime$ (for $x ≠ −0.5$).

Find the derivative of each piece of your piecewise function.
Then evaluate these limits. Do they agree?

$\lim\limits_{x\rightarrow -0.5^{-}}f'(x)= \ \ \ \ \ \text{ and }\lim \limits_{x\rightarrow -0.5^{+}}f'(x)=$

4. Does $\lim\limits _ { h \rightarrow 0 ^ { - } } \frac { f ( - 0.5 + h ) - f ( - 0.5 ) } { h } = \lim\limits _ { h \rightarrow 0 ^ { + } } \frac { f ( - 0.5 + h ) - f ( - 0.5 ) } { h }$? Justify your answer.

Notice Hana's definition of the derivative! This question is asking $f^\prime\left(−0.5\right)$ from the left agrees with $f^\prime\left(−0.5\right)$ from the right.
Looking at the graphs, do the slopes appear to agree from both sides of $−0.5$?

Use the eTool below to help solve the problem.
Click the link at right for the full version of the eTool: Calc 5-103 HW eTool