Question

Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = e−4x, [0, 2] Yes, it does not matter if f is continuous or differentiable; every function satisfies the Mean Value Theorem. No, f is not continuous on [0, 2]. There is not enough information to verify if this function satisfies the Mean Value Theorem. No, f is continuous on [0, 2] but not differentiable on (0, 2). Yes, f is continuous and differentiable on , so it is continuous on [0, 2] and differentiable on (0, 2) . If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem. (Enter your answers as a comma-separated list. If it does not satisfy the hypotheses, enter DNE).

Answer 1

Answer:

(a) Yes, it does not matter if f is continuous or differentiable; every function satisfies the Mean Value Theorem.

(b) $c =0.51995$

Step-by-step explanation:

Given

$f(x) = e^{-4x};\ [0,2]$

Solving (a); Does the function satisfy M.V.T on the given interval

We have:

$f(x) = e^{-4x};\ [0,2]$

The above function is an exponential function, and it is differentiable and continuous everywhere

Solving (b): The value of c

To do this, we use:

$f'(c) = \frac{f(b) - f(a)}{b - a}$

In this case:

$[a,b] = [0,2]$

So, we have:

$f'(c) = \frac{f(2) - f(0)}{2 - 0}$

$f'(c) = \frac{f(2) - f(0)}{2}$

Calculate f(2) and f(0)

$f(x) = e^{-4x}$

So:

$f(2) = e^{-4*2} = e^{-8} = 0.00033546262$

$f(0) = e^{-4*0} = e^{0} = 1$

This gives:

$f'(c) = \frac{0.00033546262 - 1}{2}$

$f'(c) = \frac{-0.99966453738}{2}$

$f'(c) = -0.4998$

Note that:

$f'(x) = (e^{-4x})'$

$f'(x) = -4e^{-4x}$

This implies that:

$f'(c) = -4e^{-4c}$

So, we have:

$f'(c) = -0.4998$

$-4e^{-4c} =-0.4998$

Divide both sides by -4

$e^{-4c} =\frac{-0.4998}{-4}$

$e^{-4c} =0.12495$

Take natural logarithm of both sides

$\ln(e^{-4c}) =\ln(0.12495)$

$\ln(e^{-4c}) =-2.0798$

Apply law of natural logarithm

$\ln(e^{ax}) =ax$

So:

$-4c =-2.0798$

Solve for c

$c =\frac{-2.0798}{-4}$

$c =0.51995$