# 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 $$f(a)=f(b)$$$).

Enter a function:

Enter an interval: $$[$$$, $$]$$$

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

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