mean value theorem?

Question

determine whether f satisfies the conditions of the mean value theorem on the interval [-1,2] and if so find the values of c guaranteed by the theorem.

f(x)=
2+x^3 , x 3x, x>1

(i did f(2)-f(1) over -1-2 = -7/3=3x^2.....c=sqrt7/9????)

Answer 1

The function f(x) must be continuous on a closed interval [-1,2] and differentiable on an open interval (-1,2) for the mean value theorem to apply.

f(x) is clearly continuous on [-1,2]. Additionally, its derivative exists everywhere and is:

f'(x) = 3*x^2, x <= 1 and 3 for x > 1

The MVT says that there exists a value c such that:

f'(c) = ( f(2) - f(-1) )/(2 - -1) = ( f(2) - f(-1) )/3 = ( 6 - 1 )/3 = 5/3

So all you need to do is solve:

f'(c) = 5/3

for c. Because f'(x) is constant and not equal to 5/3, for x > 1, your solution must be with x <= 1, where f'(x) = 3*x^2. So, you know that:

3*c^2 = 5/3

which means:

9*c^2 = 5

and:

c^2 = 5/9

thus,

c = +/- sqrt(5)/3

Either value of c (positive or negative) will work.

Answer 2

f(2) = 6
f(-1) = 1
[f(2) - f(-1)]/2+1 = 5/3

5/3 = f'(c)
5/3 = 3 nope, not in x >1

5/3 = 3c^2
5/9 = c^2
- √(5)/3 = c
+ √(5)/3 = c

i hope this is right, :P