Калькулятор теореми про середнє арифметичне

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