Question

consider the following function.
$f(x)=1 - x^{2/3}$
find $f(-1)$ and $f(1)$.
$f(-1)=$
$f(1)=$
find all values $c$ in $(-1,1)$ such that $f(c)=0$. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.)
$c=$
based off of this information, what conclusions can be made about rolles theorem?
$\bigcirc$ this contradicts rolles theorem, since $f$ is differentiable, $f(-1)=f(1)$, and $f(c)=0$ exists, but $c$ is not in $(-1,1)$.
$\bigcirc$ this does not contradict rolles theorem, since $f(0)=0$, and $0$ is in the interval $(-1,1)$.
$\bigcirc$ this contradicts rolles theorem, since $f(-1)=f(1)$, there should exist a number $c$ in $(-1,1)$ such that $f(c)=0$.
$\bigcirc$ this does not contradict rolles theorem, since $f(0)$ does not exist, and so $f$ is not differentiable on $(-1,1)$.
$\bigcirc$ nothing can be concluded.

Explanation:

Step1: Calculate \( f(-1) \) and \( f(1) \)

Substitute \( x=-1 \) into \( f(x)=1 - x^{\frac{2}{3}} \):
\( f(-1)=1-(-1)^{\frac{2}{3}}=1 - 1 = 0 \)
Substitute \( x = 1 \) into \( f(x)=1 - x^{\frac{2}{3}} \):
\( f(1)=1-(1)^{\frac{2}{3}}=1 - 1 = 0 \)

Step2: Find the derivative \( f^{\prime}(x) \)

Using the power rule \( (x^n)^\prime=nx^{n - 1} \), \( f(x)=1 - x^{\frac{2}{3}} \), then \( f^{\prime}(x)=-\frac{2}{3}x^{-\frac{1}{3}}=-\frac{2}{3\sqrt[3]{x}} \)

Step3: Solve \( f^{\prime}(c) = 0 \)

Set \( f^{\prime}(c)=-\frac{2}{3\sqrt[3]{c}} = 0 \). The equation \( -\frac{2}{3\sqrt[3]{c}}=0 \) has no solution since the numerator \(- 2
eq0\) for all real \( c
eq0 \).

Step4: Analyze Rolle's Theorem conditions

Rolle's Theorem requires \( f(x) \) to be continuous on \([-1,1]\), \( f(x) \) is continuous on \([-1,1]\) (since \( y = x^{\frac{2}{3}}=\sqrt[3]{x^{2}}\) is continuous for all real \(x\)). \( f(-1)=f(1) = 0 \). But \( f^{\prime}(x)=-\frac{2}{3\sqrt[3]{x}}\) is not defined at \(x = 0\), so \(f(x)\) is not differentiable on \((-1,1)\) (because \(x = 0\in(-1,1)\) and \(f^{\prime}(0)\) does not exist).

Answer:

\( f(-1)=0 \), \( f(1)=0 \), \( c=\text{DNE} \), and the correct option is "This does not contradict Rolle's Theorem, since \( f^{\prime}(0) \) does not exist, and so \( f \) is not differentiable on \((-1,1)\)."