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agasfer [191]
3 years ago
15

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 m

atter 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).
Mathematics
1 answer:
Zigmanuir [339]3 years ago
3 0

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

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