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?
this contradicts rolles theorem, since f is differentiable, f(-1)=f(1), and f(c)=0 exists, but c is not in (-1, 1).
this does not contradict rolles theorem, since f(0)=0, and 0 is in the interval (-1, 1).
this contradicts rolles theorem, since f(-1)=f(1), there should exist a number c in (-1, 1) such that f(c)=0.
this contradicts rolles theorem, since f(-1)=f(1), there should exist a number c in (-1, 1) such that f(c)=0.
this does not contradict rolles theorem, since f(0) does not exist, and so f is not differentiable on (-1, 1).
nothing can be concluded.

Explanation:

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

Substitute \(x = - 1\) into \(f(x)=1 - x^{2/3}\).
\[

$$\begin{align*} f(-1)&=1-(-1)^{2/3}\\ &=1 - 1\\ &=0 \end{align*}$$

\]

Step2: Calculate \(f(1)\)

Substitute \(x = 1\) into \(f(x)=1 - x^{2/3}\).
\[

$$\begin{align*} f(1)&=1-1^{2/3}\\ &=1 - 1\\ &=0 \end{align*}$$

\]

Step3: Find the derivative of \(f(x)\)

Use the power - rule \((x^n)^\prime=nx^{n - 1}\).
\[

$$\begin{align*} f(x)&=1 - x^{2/3}\\ f^\prime(x)&=0-\frac{2}{3}x^{2/3 - 1}\\ &=-\frac{2}{3}x^{-1/3}=-\frac{2}{3\sqrt[3]{x}} \end{align*}$$

\]
Set \(f^\prime(c) = 0\), \(-\frac{2}{3\sqrt[3]{c}}=0\). There is no value of \(c\) for which \(-\frac{2}{3\sqrt[3]{c}} = 0\) since the numerator is non - zero. So \(c=\text{DNE}\).

Step4: Analyze Rolle's Theorem

Rolle's Theorem states that if \(y = f(x)\) is continuous on \([a,b]\), differentiable on \((a,b)\) and \(f(a)=f(b)\), then there exists at least one \(c\in(a,b)\) such that \(f^\prime(c)=0\). Here \(a=-1\), \(b = 1\), \(f(-1)=f(1) = 0\), but \(f(x)=1 - x^{2/3}\) is not differentiable at \(x = 0\in(-1,1)\) because \(f^\prime(x)=-\frac{2}{3\sqrt[3]{x}}\) is undefined at \(x = 0\). So this does not contradict Rolle's Theorem, since \(f^\prime(0)\) does not exist, and so \(f\) is not differentiable on \((-1,1)\).

Answer:

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