At what x value on the interval [-2,3] does the graph of f(x)=x^2+2x-1 satisfy the mean value theorem?

The Mean Value Theorem states that if a function is
continuous and differentiable over the closed interval [a,b], then there is a point c,
that belongs to [a,b], such that:

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

To calculate f'(c), we'll have to differentiate
the function f(x).

f'(x) =
(x^2+2x-1)'

f'(x) = 2x +
2

Now, we'll substitute x by
c:

f'(c) = 2c + 2

From the
mean value theorem, we'll get:

f'(c) = [f(3) - f(-2)]/(3 +
2)

We'll calculate f(3):

f(3)
= 3^2+2*3-1

f(3) =9+6-1

f(3)
=14

We'll calculate
f(-2):

f(-2) =
(-2)^2+2*(-2)-1

f(-2) = 4 - 4 -
1

We'll eliminate like
terms:

f(-2) = -1

f'(c) =
(14+1)/5

f'(c) = 15/5

f'(c) =
3 (1)

But f'(c) = 2c + 2
(2).

We'll substitute (1) in
(2):

2c + 2 = 3

2c =
3-2

2c = 1

c =
1/2

For x = 1/2, the mean
value theorem is satisfied for the function f(x) =
x^2+2x-1.