Question

Suppose that f(0)=3 and f′(x)≤7 for all values of x. Use the Mean Value Theorem to determine how large f(4) can possibly be.

Answer 1

With these given values:

 f(0)=4 and f′(x)≤4 

The solution is as follows:

 f(0) = 4 
f ' (x) ≤ 4 for all x 

The Mean Value Theorem says there is a c between 0 and 4 such that f ' (c) = [f(4) - f(0)] / (4 - 0) 
But since f ' (x) ≤ 4 for all x, then [f(4) - f(0)] / (4 - 0) ≤ 4 
Since f(0) = 4, then [f(4) - 4] / (4 - 0) ≤ 4 
Multiplying both sides by 4 you get f(4) - 4 ≤ 16 
Adding 4 to both sides you get f(4) ≤ 20 

So f(4) ≤ 20

By examining the solution, it could guide you on answering the problem on your own. Hope that helps

Answer 2

Mean value theorem state that the f'(c) is equal to the function's average rate of change over [a, b].

The maximum value of f(4) will be 31.

Mean value theorem:

It states that if a function f is continuous on the closed interval [a ,b] and differentiable on the open interval (a ,b), then there exists a point c in the interval (a, b) such that f'(c) is equal to the function's average rate of change over [a, b].

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

It is given that,  f(0)=3

                       $f'(x)=\frac{f(4)-f(0)}{4-0}$

Therefore,    

                          $f'(x)\leq 7\\\\\frac{f(4)-f(0)}{4}\leq 7\\ \\\frac{f(4)-3}{4}\leq 7\\\\f(4)\leq 28+3\\\\f(4)\leq 31$

Hence, the maximum value of f(4) will be 31.

Learn more about the differentiation here:

https://brainly.com/question/954654