contestada

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.

Respuesta :

kudzordzifrancis

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

[tex]f(x)=16-\frac{x^2}{3}[/tex]

To find [tex]f(-64)[/tex], we substitute [tex]x=-64[/tex] into the function.

[tex]f(-64)=16-\frac{(-64)^2}{3}[/tex]

[tex]f(-64)=16-\frac{4096}{3}[/tex]

[tex]f(-64)=-\frac{4048}{3}[/tex]

To find [tex]f(64)[/tex], we substitute [tex]x=64[/tex] into the function.

[tex]f(64)=16-\frac{(64)^2}{3}[/tex]

[tex]f(64)=16-\frac{4096}{3}[/tex]

[tex]f(64)=-\frac{4048}{3}[/tex]

To find [tex]f'(c)[/tex], we must first find [tex]f'(x)[/tex].

[tex]f'(x)=-\frac{2x}{3}[/tex]

This implies that;

[tex]f'(c)=-\frac{2c}{3}[/tex]

[tex]f'(c)=0[/tex]

[tex]\Rightarrow -\frac{2c}{3}=0[/tex]

[tex]\Rightarrow -\frac{2c}{3}\times -\frac{3}{2}=0\times -\frac{3}{2}[/tex]

[tex]c=0[/tex]

For this function to satisfy the Rolle's Theorem;

It must be continuous on [tex][-64,64][/tex].

It must be differentiable  on [tex](-64,64)[/tex].

and

[tex]f(-64)=f(64)[/tex].

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:

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

[tex]f(x)=16-x^{\frac{2}{3}}[/tex]

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,

[tex]f(64)=16-(64)^{\frac{2}{3}}=0[/tex]

At x=-64,

[tex]f(-64)=16-(-64)^{\frac{2}{3}}=0[/tex]

So, f(−64) = f(64).

Differentiate the given function with respect to x.

[tex]f'(x)=0-\frac{2}{3}x^{-\frac{1}{3}}[/tex]

[tex]f'(x)=-\frac{2}{3x^{\frac{1}{3}}}[/tex]

Substitute x=c,

[tex]f'(c)=-\frac{2}{3c^{\frac{1}{3}}}[/tex]

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

[tex]f'(c)=0[/tex]

[tex]-\frac{2}{3c^{\frac{1}{3}}}=0[/tex]

[tex]-2=0[/tex]

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.

Otras preguntas