# Mean Value Theorem

• May 29th 2011, 05:26 AM
Jon123
Mean Value Theorem
Hello!

I need some help please with two revision questions.
I read my Calculus book and I did not understand it!

Q1.
Suppose that 3<= f ' (x) <= 5 for all values of x. Determine the highest and lowest possible values for the expression A = f (8) - f (2).

Am I meant to use "MVT" ??

Q2.
For what values of the numbers a and b does the function f(x) = axe^bx²
have the maximum value f(2) = 1? Give reasons.

Sorry this I do not know.
• May 29th 2011, 06:21 AM
Also sprach Zarathustra
f'(x)={f(x)-f(y)}/{x-y}

for the second one:

put (2,1) into f(x).

this is your first equation with a and b.

now,
find f'(x).

f'(x)=0

put x=1.

And there you have second equation.

solve system of equations finding a and b.
• May 29th 2011, 06:29 AM
HallsofIvy
Quote:

Originally Posted by Jon123
Hello!

I need some help please with two revision questions.
I read my Calculus book and I did not understand it!

Q1.
Suppose that 3<= f ' (x) <= 5 for all values of x. Determine the highest and lowest possible values for the expression A = f (8) - f (2).

Am I meant to use "MVT" ??

That seems most reasonable to me (f(8)- f(2))/(8- 2)= (f(8)- f(2))/6 is equal to the derivative of f at some point- but you know that can never be larger than 5 or less than 3.

Quote:

Q2.
For what values of the numbers a and b does the function f(x) = axe^bx²
have the maximum value f(2) = 1? Give reasons.

At a max or min, the derivative will be 0. $f'(x)= ae^{bx}+ abxe^{bx}$. Where is that equal to 0? What is the value of f there? What must a and b equal in order that that result be 1?

Quote:

Sorry this I do not know.