# Mean Value Theorem for Continuous Functions

## Key Questions

• 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.

• Actually, Rolle's Theorem require differentiablity, and it is a special case of Mean Value Theorem.
Please watch this video for more details.

• Mean Value Theorem
If a function $f$ is continuous on $\left[a , b\right]$ and differentiable on $\left(a , b\right)$,
then there exists c in $\left(a , b\right)$ such that $f ' \left(c\right) = \frac{f \left(b\right) - f \left(a\right)}{b - a}$.