# Thread: Rolle Theorem and MVT?

1. ## Rolle Theorem and MVT?

In the interval [0,3] f(x)= {1 , x=0
ax+b , 0 less than x less than or equal to 1
x2+4x+c , 1 less than x less than or equal to 3
(if it isnt clear, f(x) is a piecewise function)

How do i solve for a, b and c?
thanks!

2. you didn't say so, so let me guess ... you need to find the values of a, b, and c so that the MVT will apply to f(x) on the interval [0,3] ?

If the MVT applies, then f(x) must satisfy what two conditions?

3. Sorry for not making it clear!

I have to determine the values of a,b and c such that the function f satisfies the hypotheses of the Mean Value Theorem on the interval [0,3]
Does that make more sense?

4. f(x) needs to be continuous on [0,3] and differentiable on (0,3).

$\displaystyle \lim_{x \to 0^+} ax + b = 1$

so , that means $\displaystyle b = 1$

to be continuous at x = 1 ...

$\displaystyle f(1) = a+1$

$\displaystyle \lim_{x \to 1^-} ax+1 = \lim_{x \to 1^+} x^2 + 4x + c$

$\displaystyle a+1 = 5+c$

to be differentiable at x = 1 ...

$\displaystyle \lim_{x \to 1^-} \frac{(ax+1) - (a+1)}{x - 1}$

$\displaystyle \lim_{x \to 1^-} \frac{a(x-1)}{x - 1} = a$

$\displaystyle \lim_{x \to 1^+} \frac{(x^2+4x+c) - (5+c)}{x - 1}$

$\displaystyle \lim_{x \to 1^+} \frac{x^2+4x-5}{x - 1}$

$\displaystyle \lim_{x \to 1^+} \frac{(x+5)(x-1)}{x - 1} = 6$

$\displaystyle a = 6$

$\displaystyle c = 2$

5. Thanks so much. I appreciate it!