# Calculadora del teorema del valor medio

La calculadora encontrará todos los números $c$ (con los pasos mostrados) que satisfagan las conclusiones del teorema del valor medio para la función dada en el intervalo dado. El teorema de Rolle es un caso especial del teorema del valor medio (cuando $f (a) = f (b)$).

Enter a function:

Enter an interval: $[$, $]$

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.

## 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 the interval $(a,b)$, 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: $- \frac{10 \sqrt{3}}{3}\approx -5.77350269189626$, $\frac{10 \sqrt{3}}{3}\approx 5.77350269189626$