Question

Consider the following function. f(x) = 16 − x2/3 Find f(−64) and f(64). f(−64) = f(64) = Find all values c in (−64, 64) 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 Rolle's Theorem? This contradicts Rolle's Theorem, since f is differentiable, f(−64) = f(64), and f '(c) = 0 exists, but c is not in (−64, 64). This does not contradict Rolle's Theorem, since f '(0) = 0, and 0 is in the interval (−64, 64). This contradicts Rolle's Theorem, since f(−64) = f(64), there should exist a number c in (−64, 64) such that f '(c) = 0. This does not contradict Rolle's Theorem, since f '(0) does not exist, and so f is not differentiable on (−64, 64). Nothing can be concluded.

Answer 1

Answer:

This does not contradict Rolle's Theorem, since f '(0) = 0, and 0 is in the interval (−64, 64).

Step-by-step explanation:

The given function is

$f(x)=16-\frac{x^2}{3}$

To find $f(-64)$, we substitute $x=-64$ into the function.

$f(-64)=16-\frac{(-64)^2}{3}$

$f(-64)=16-\frac{4096}{3}$

$f(-64)=-\frac{4048}{3}$

To find $f(64)$, we substitute $x=64$ into the function.

$f(64)=16-\frac{(64)^2}{3}$

$f(64)=16-\frac{4096}{3}$

$f(64)=-\frac{4048}{3}$

To find $f'(c)$, we must first find $f'(x)$.

$f'(x)=-\frac{2x}{3}$

This implies that;

$f'(c)=-\frac{2c}{3}$

$f'(c)=0$

$\Rightarrow -\frac{2c}{3}=0$

$\Rightarrow -\frac{2c}{3}\times -\frac{3}{2}=0\times -\frac{3}{2}$

$c=0$

For this function to satisfy the Rolle's Theorem;

It must be continuous on $[-64,64]$.

It must be differentiable  on $(-64,64)$.

and

$f(-64)=f(64)$.

All the hypotheses are met, hence this does not contradict Rolle's Theorem, since f '(0) = 0, and 0 is in the interval (−64, 64) is the correct choice.

Answer 2

Answer:

c = DNE

Option C: This contradicts Rolle's Theorem, since f(−64) = f(64), there should exist a number c in (−64, 64) such that f '(c) = 0.

Step-by-step explanation:

The given function is

$f(x)=16-x^{\frac{2}{3}}$

Rolle's theorem states that if the function f is

1. Continuous on [a, b],

2. Differentiable on the open interval (a, b) such that f(a) = f(b),

then f′(x) = 0 for some x with a ≤ x ≤ b.

At x=64,

$f(64)=16-(64)^{\frac{2}{3}}=0$

At x=-64,

$f(-64)=16-(-64)^{\frac{2}{3}}=0$

So, f(−64) = f(64).

Differentiate the given function with respect to x.

$f'(x)=0-\frac{2}{3}x^{-\frac{1}{3}}$

$f'(x)=-\frac{2}{3x^{\frac{1}{3}}}$

Substitute x=c,

$f'(c)=-\frac{2}{3c^{\frac{1}{3}}}$

We need to find the value of c such that f '(c) = 0.

$f'(c)=0$

$-\frac{2}{3c^{\frac{1}{3}}}=0$

$-2=0$

This equation is not true for any value of c. So, the value of does not exist.

This contradicts Rolle's Theorem, since f(−64) = f(64), there should exist a number c in (−64, 64) such that f '(c) = 0.

Therefore, the correct option is C.