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

consider the following function.\n\nf(x)=1 - x^{2/3}\n\nfind f(-1) and f(1).\nf(-1)=\n\nf(1)=\n\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.)\nc=\n\nbased off of this information, what conclusions can be made about rolles theorem?\nthis contradicts rolles theorem, since f is differentiable, f(-1)=f(1), and f(c)=0 exists, but c is not in (-1, 1).\nthis does not contradict rolles theorem, since f(0)=0, and 0 is in the interval (-1, 1).\nthis contradicts rolles theorem, since f(-1)=f(1), there should exist a number c in (-1, 1) such that f(c)=0.\nthis contradicts rolles theorem, since f(-1)=f(1), there should exist a number c in (-1, 1) such that f(c)=0.\nthis does not contradict rolles theorem, since f(0) does not exist, and so f is not differentiable on (-1, 1).\nnothing can be concluded.

consider the following function.\n\nf(x)=1 - x^{2/3}\n\nfind f(-1) and f(1).\nf(-1)=\n\nf(1)=\n\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.)\nc=\n\nbased off of this information, what conclusions can be made about rolles theorem?\nthis contradicts rolles theorem, since f is differentiable, f(-1)=f(1), and f(c)=0 exists, but c is not in (-1, 1).\nthis does not contradict rolles theorem, since f(0)=0, and 0 is in the interval (-1, 1).\nthis contradicts rolles theorem, since f(-1)=f(1), there should exist a number c in (-1, 1) such that f(c)=0.\nthis contradicts rolles theorem, since f(-1)=f(1), there should exist a number c in (-1, 1) such that f(c)=0.\nthis does not contradict rolles theorem, since f(0) does not exist, and so f is not differentiable on (-1, 1).\nnothing can be concluded.

Answer

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))