Answer:
(a) Yes, it does not matter if f is continuous or differentiable; every function satisfies the Mean Value Theorem.
(b) 
Step-by-step explanation:
Given
![f(x) = e^{-4x};\ [0,2]](https://tex.z-dn.net/?f=f%28x%29%20%3D%20e%5E%7B-4x%7D%3B%5C%20%5B0%2C2%5D)
Solving (a); Does the function satisfy M.V.T on the given interval
We have:
![f(x) = e^{-4x};\ [0,2]](https://tex.z-dn.net/?f=f%28x%29%20%3D%20e%5E%7B-4x%7D%3B%5C%20%5B0%2C2%5D)
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:

In this case:
![[a,b] = [0,2]](https://tex.z-dn.net/?f=%5Ba%2Cb%5D%20%3D%20%5B0%2C2%5D)
So, we have:


Calculate f(2) and f(0)

So:


This gives:



Note that:


This implies that:

So, we have:


Divide both sides by -4


Take natural logarithm of both sides


Apply law of natural logarithm

So:

Solve for c

Answer:
x² - 6x + 9
Step-by-step explanation:
Answer:
-----------------------------------------
Given:
- Function f(x) = 2|x + 3| - 2
Find a function g which is a horizontal shrink by a factor of 1/3 of f(x).
Horizontal shrink by a factor of k is:
Apply this to the given function:
- g(x) = f(x : 1/3) = f(3x) = 2|3x + 3| - 2 = 6|x + 1| - 2
C. If the first coefficient is negative, then the graph opens downwards
Pop ok you see so the answer here would be frantically elastic