# Satisfying the Mean Value Theorem

## Homework Statement

For what values of a,m, and b does the function

{3 ,x=0
f(x)= {-x^2 + 3x +a ,0<x<1
{mx + b ,1<=x<=2

satisfy the hypothesis of the Mean Value Theorem of the interval [0,2].

## Homework Equations

(f(b) - f(a))/(b-a) = f'(c)

## The Attempt at a Solution

So, I wanted to make a point of continuity for the whole equation. So, I set the equation to x=1 in both cases and then equated them, and got;
2+a=m+b ---> a=m+b-2

Then, I found the derivative of the original equation and set x again to 1, then equated the two given that were diffable, and got;
m+b=1+a

Now I can't get any values for a,b, or m without getting:
3=0

I must be doing something wrong.

LCKurtz
Homework Helper
Gold Member

## Homework Statement

For what values of a,m, and b does the function

satisfy the hypothesis of the Mean Value Theorem of the interval [0,2].

## Homework Equations

(f(b) - f(a))/(b-a) = f'(c)

## The Attempt at a Solution

So, I wanted to make a point of continuity for the whole equation. So, I set the equation to x=1 in both cases and then equated them, and got;
2+a=m+b ---> a=m+b-2

Then, I found the derivative of the original equation and set x again to 1, then equated the two given that were diffable, and got;
m+b=1+a

The derivative of mx + b is not m + b and the derivative of -x2+3x + a is not -2x+3+a. Also don't forget you will need continuity at x = 0.