Answer:
(a) Yes, it does not matter if f is continuous or differentiable; every function satisfies the Mean Value Theorem.
(b) [tex]c =0.51995[/tex]
Step-by-step explanation:
Given
[tex]f(x) = e^{-4x};\ [0,2][/tex]
Solving (a); Does the function satisfy M.V.T on the given interval
We have:
[tex]f(x) = e^{-4x};\ [0,2][/tex]
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:
[tex]f'(c) = \frac{f(b) - f(a)}{b - a}[/tex]
In this case:
[tex][a,b] = [0,2][/tex]
So, we have:
[tex]f'(c) = \frac{f(2) - f(0)}{2 - 0}[/tex]
[tex]f'(c) = \frac{f(2) - f(0)}{2}[/tex]
Calculate f(2) and f(0)
[tex]f(x) = e^{-4x}[/tex]
So:
[tex]f(2) = e^{-4*2} = e^{-8} = 0.00033546262[/tex]
[tex]f(0) = e^{-4*0} = e^{0} = 1[/tex]
This gives:
[tex]f'(c) = \frac{0.00033546262 - 1}{2}[/tex]
[tex]f'(c) = \frac{-0.99966453738}{2}[/tex]
[tex]f'(c) = -0.4998[/tex]
Note that:
[tex]f'(x) = (e^{-4x})'[/tex]
[tex]f'(x) = -4e^{-4x}[/tex]
This implies that:
[tex]f'(c) = -4e^{-4c}[/tex]
So, we have:
[tex]f'(c) = -0.4998[/tex]
[tex]-4e^{-4c} =-0.4998[/tex]
Divide both sides by -4
[tex]e^{-4c} =\frac{-0.4998}{-4}[/tex]
[tex]e^{-4c} =0.12495[/tex]
Take natural logarithm of both sides
[tex]\ln(e^{-4c}) =\ln(0.12495)[/tex]
[tex]\ln(e^{-4c}) =-2.0798[/tex]
Apply law of natural logarithm
[tex]\ln(e^{ax}) =ax[/tex]
So:
[tex]-4c =-2.0798[/tex]
Solve for c
[tex]c =\frac{-2.0798}{-4}[/tex]
[tex]c =0.51995[/tex]
