consider the following function.\n$f(x)=1 - x^{2/3}$\nfind $f(-1)$ and $f(1)$.\n$f(-1)=$\n$f(1)=$\nfind all…

consider the following function.\n$f(x)=1 - x^{2/3}$\nfind $f(-1)$ and $f(1)$.\n$f(-1)=$\n$f(1)=$\nfind 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.)\n$c=$\nbased off of this information, what conclusions can be made about rolles theorem?\n$\\bigcirc$ this contradicts rolles theorem, since $f$ is differentiable, $f(-1)=f(1)$, and $f(c)=0$ exists, but $c$ is not in $(-1,1)$.\n$\\bigcirc$ this does not contradict rolles theorem, since $f(0)=0$, and $0$ is in the interval $(-1,1)$.\n$\\bigcirc$ this contradicts rolles theorem, since $f(-1)=f(1)$, there should exist a number $c$ in $(-1,1)$ such that $f(c)=0$.\n$\\bigcirc$ this does not contradict rolles theorem, since $f(0)$ does not exist, and so $f$ is not differentiable on $(-1,1)$.\n$\\bigcirc$ nothing can be concluded.

consider the following function.\n$f(x)=1 - x^{2/3}$\nfind $f(-1)$ and $f(1)$.\n$f(-1)=$\n$f(1)=$\nfind 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.)\n$c=$\nbased off of this information, what conclusions can be made about rolles theorem?\n$\\bigcirc$ this contradicts rolles theorem, since $f$ is differentiable, $f(-1)=f(1)$, and $f(c)=0$ exists, but $c$ is not in $(-1,1)$.\n$\\bigcirc$ this does not contradict rolles theorem, since $f(0)=0$, and $0$ is in the interval $(-1,1)$.\n$\\bigcirc$ this contradicts rolles theorem, since $f(-1)=f(1)$, there should exist a number $c$ in $(-1,1)$ such that $f(c)=0$.\n$\\bigcirc$ this does not contradict rolles theorem, since $f(0)$ does not exist, and so $f$ is not differentiable on $(-1,1)$.\n$\\bigcirc$ nothing can be concluded.

Answer

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\neq0) for all real ( c\neq0 ).

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))."