# Mean value theorem problem

• Saitama
In summary, the author attempted to solve a homework equation but was unsuccessful because the c value could take on any value greater than -3.

## Homework Statement

Suppose that f(0)=-3 and f'(x)<=5 for all values of x. The the largest value of f(2) is
A)7
B)-7
C)13
D)8

## The Attempt at a Solution

The problem can be easily solved using the mean value theorem but solving it in a different way doesn't give the right answer and I am not sure if the following is a valid approach.
I have ##f'(x) \leq 5 \Rightarrow f(x) \leq 5x+c##, where c is some constant. At x=0, ##f(0) \leq c \Rightarrow c\geq -3##.
At x=2, ##f(2)\leq 10+c##. The problem is that c can take any value greater than -3 and due to this I reach no answer. What is wrong with this method?

Any help is appreciated. Thanks!

Pranav-Arora said:

## Homework Statement

Suppose that f(0)=-3 and f'(x)<=5 for all values of x. The the largest value of f(2) is
A)7
B)-7
C)13
D)8

## The Attempt at a Solution

The problem can be easily solved using the mean value theorem but solving it in a different way doesn't give the right answer and I am not sure if the following is a valid approach.
I have ##f'(x) \leq 5 \Rightarrow f(x) \leq 5x+c##, where c is some constant. At x=0, ##f(0) \leq c \Rightarrow c\geq -3##.
At x=2, ##f(2)\leq 10+c##. The problem is that c can take any value greater than -3 and due to this I reach no answer. What is wrong with this method?

Any help is appreciated. Thanks!

f ' (x) = k ≤ 5. Integrating, f=kx+c and f(0)=-3. Therefore C=-3, well determined.

ehild

ehild said:
f ' (x) = k ≤ 5. Integrating, f=kx+c and f(0)=-3. Therefore C=-3, well determined.

ehild

Really silly on my part, thank you very much ehild! :)