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Exercise 17Calculus › Single-variable calculus › Differentiation › Mean Value TheoremHigh schoolAdded 4 Nov 2026

The mean value theorem

#mean value theorem#Lagrange#existence

Statement

Let f(x)=x2f(x)=x^2 on [0,2][0,2]. Find the point cc guaranteed by the mean value theorem.

Hint

Solve f′(c)=f(2)−f(0)2−0f'(c)=\frac{f(2)-f(0)}{2-0}.

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Solution

Right side. f(2)−f(0)2=4−02=2\frac{f(2)-f(0)}{2}=\frac{4-0}{2}=2.

Derivative. f′(c)=2cf'(c)=2c. Solve 2c=22c=2, so c=1c=1.

Conclusion. c=1∈]0,2[c=1\in]0,2[.

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