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

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### 1 Answer

We'll recall the identity established by the Mean Value Theorem:

f'(c) = [f(b) - f(a)]/(b-a), c belongs to the interval [a,b].

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

**If c = 1/2, the mean value theorem is verified for the function f(x) = x^2+2x-1.**