# How do I find the numbers c that satisfy the Mean Value Theorem for f(x)=x^3+x-1 on the interval [0,3] ?

Sep 28, 2014

The value of $c$ is $\sqrt{3}$.

Let us look at some details.

M.V.Thm. states that there exists $c$ in (0,3) such that

$f ' \left(c\right) = \frac{f \left(3\right) - f \left(0\right)}{3 - 0}$.

Let us find such $c$.

The left-hand side is

$f ' \left(c\right) = 3 {c}^{2} + 1$.

The right-hand side is

$\frac{f \left(3\right) - f \left(0\right)}{3 - 0} = \frac{29 - \left(- 1\right)}{3} = 10$.

By setting them equal to each other,

3c^2+1=10 Rightarrow 3x^2=9 Rightarrow x^2=3 Rightarrow x=pm sqrt{3}

Since $0 < c < 3$, $c = \sqrt{3}$.

I hope that this was helpful.