consider the following function.\n\n$f(x)=4 - x^{2/3}$\n\nfind $f(-8)$ and $f(8)$.\n\n$f(-8)=$\n\n$f(8)=$\n\n…

consider the following function.\n\n$f(x)=4 - x^{2/3}$\n\nfind $f(-8)$ and $f(8)$.\n\n$f(-8)=$\n\n$f(8)=$\n\nfind all values $c$ in $(-8, 8)$ such that $f(c)=0$. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.)\n\n$c=$\n\nbased off of this information, what conclusions can be made about rolles theorem?\n\n$\bigcirc$ this contradicts rolles theorem, since $f$ is differentiable, $f(-8)=f(8)$, and $f(c)=0$ exists, but $c$ is not in $(-8,8)$.\n\n$\bigcirc$ this does not contradict rolles theorem, since $f(0)=0$, and 0 is in the interval $(-8,8)$.\n\n$\bigcirc$ this contradicts rolles theorem, since $f(-8)=f(8)$, there should exist a number $c$ in $(-8,8)$ such that $f(c)=0$.\n\n$\bigcirc$ this does not contradict rolles theorem, since $f(0)$ does not exist, and so $f$ is not differentiable on $(-8,8)$.\n\n$\bigcirc$ nothing can be concluded.

consider the following function.\n\n$f(x)=4 - x^{2/3}$\n\nfind $f(-8)$ and $f(8)$.\n\n$f(-8)=$\n\n$f(8)=$\n\nfind all values $c$ in $(-8, 8)$ such that $f(c)=0$. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.)\n\n$c=$\n\nbased off of this information, what conclusions can be made about rolles theorem?\n\n$\bigcirc$ this contradicts rolles theorem, since $f$ is differentiable, $f(-8)=f(8)$, and $f(c)=0$ exists, but $c$ is not in $(-8,8)$.\n\n$\bigcirc$ this does not contradict rolles theorem, since $f(0)=0$, and 0 is in the interval $(-8,8)$.\n\n$\bigcirc$ this contradicts rolles theorem, since $f(-8)=f(8)$, there should exist a number $c$ in $(-8,8)$ such that $f(c)=0$.\n\n$\bigcirc$ this does not contradict rolles theorem, since $f(0)$ does not exist, and so $f$ is not differentiable on $(-8,8)$.\n\n$\bigcirc$ nothing can be concluded.

Answer

Explanation:

Step1: Calculate ( f(-8) )

Substitute ( x = -8 ) into ( f(x)=4 - x^{2/3} ). ( f(-8)=4-(-8)^{2/3}=4 - [(-2)^3]^{2/3}=4 - (-2)^2=4 - 4=0 )

Step2: Calculate ( f(8) )

Substitute ( x = 8 ) into ( f(x)=4 - x^{2/3} ). ( f(8)=4 - 8^{2/3}=4 - (2^3)^{2/3}=4 - 2^2=4 - 4=0 )

Step3: Find the derivative ( f'(x) )

Use the power rule ( (x^n)'=nx^{n - 1} ). ( f'(x)=-\frac{2}{3}x^{-1/3}=-\frac{2}{3\sqrt[3]{x}} ) Set ( f'(c) = 0 ), ( -\frac{2}{3\sqrt[3]{c}}=0 ). This equation has no solution for ( c\in(-8,8) ) (since the numerator is non - zero). Also, ( f(x) ) is not differentiable at ( x = 0 ) (because ( f'(x)=-\frac{2}{3\sqrt[3]{x}} ) has a vertical asymptote at ( x = 0 )).

Rolle's Theorem requires that ( y = f(x) ) is continuous on ( [a,b] ), differentiable on ( (a,b) ), and ( f(a)=f(b) ). Here ( f(-8)=f(8) ), but ( f(x) ) is not differentiable on ( (-8,8) ) (due to non - differentiability at ( x = 0 )).

Answer:

( f(-8)=0 ) ( f(8)=0 ) ( c=\text{DNE} ) The correct option is: This does not contradict Rolle's Theorem, since ( f'(0) ) does not exist, and so ( f ) is not differentiable on ( (-8,8) ).