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.

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

To find , we substitute into the function.

To find , we substitute into the function.

To find , we must first find .

This implies that;

For this function to satisfy the Rolle's Theorem;

It must be continuous on .

It must be differentiable  on .

and

.

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.

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

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,

At x=-64,

So, f(−64) = f(64).

Differentiate the given function with respect to x.

Substitute x=c,

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

it is probably the graph on the bottom right.

(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)