Mathematics, 18.10.2019 14:10 Cupcake8189

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.

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Answer from: jake9919

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 from: vinp190p9zekn

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.

Answer from: Quest

answer: a video game has three difficulty levels: easy, medium, and expert. the game's scoring works in such a way that a player has to score a fixed number of points to advance to the next level, and then the scoring starts over. the total score needed to clear all three difficulty levels is 7,500. two times the number of points needed to clear the easy level is 500 more than the number of points needed to clear the medium level. twice the number of points needed to clear the medium difficulty level is 1,500 more than the number of points needed to clear the expert level. if x, y, and z represent the easy, medium, and expert levels, respectively, and a is the coefficient matrix of the system of equations modeling this situation, identify the inverse matrix, a-1, and the solution matrix, x, of the system of equations.

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Answer from: Quest

3.64 deaths an hour.