# Mean Value Theorem Calculator

The calculator will find all numbers $$$c$$$ (with steps shown) that satisfy the conclusions of the Mean Value Theorem for the given function on the given interval. Rolle's theorem is a special case of the mean value theorem (when $$$f(a)=f(b)$$$).

## Solution

**Your input: find all numbers $$$c$$$ (with steps shown) to satisfy the conclusions of the Mean Value Theorem for the function $$$f=x^{3} - 2 x$$$ on the interval $$$\left[-10, 10\right]$$$.**

The Mean Value Theorem states that for a continuous and differentiable function $$$f(x)$$$ on the interval $$$[a,b]$$$ there exists such number $$$c$$$ from that interval, that $$$f'(c)=\frac{f(b)-f(a)}{b-a}$$$.

First, evaluate the function at the endpoints of the interval:

$$$f \left( 10 \right) = 980$$$

$$$f \left( -10 \right) = -980$$$

Next, find the derivative: $$$f'(c)=3 c^{2} - 2$$$ (for steps, see derivative calculator).

Form the equation: $$$3 c^{2} - 2=\frac{\left( 980\right)-\left( -980\right)}{\left( 10\right)-\left( -10\right)}$$$

Simplify: $$$3 c^{2} - 2=98$$$

Solve the equation on the given interval: $$$c=- \frac{10 \sqrt{3}}{3}$$$, $$$c=\frac{10 \sqrt{3}}{3}$$$

**Answer: $$$- \frac{10 \sqrt{3}}{3}\approx -5.77350269189626$$$, $$$\frac{10 \sqrt{3}}{3}\approx 5.77350269189626$$$**