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Exercise 43Calculus › Single-variable calculus › Differentiation › Mean Value TheoremHigh schoolAdded 4 Nov 2026

Mean value theorem for $f(x)=x^3$

#mean value theorem#cubic#existence

Statement

Apply the mean value theorem to f(x)=x3f(x)=x^3 on [1,3][1,3] and find the point cc.

Hint

f′(c)=3c2f'(c)=3c^2; solve 3c2=27−13−13c^2=\frac{27-1}{3-1}.

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Solution

Slope. 27−12=13\frac{27-1}{2}=13.

Equation. 3c2=13⇒c=13/3≈2.083c^2=13\Rightarrow c=\sqrt{13/3}\approx2.08.

Conclusion. c=13/3∈]1,3[c=\sqrt{13/3}\in]1,3[.

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