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Proof of Mean Value Theorem in One Dimension

theoremthm:calc-mean-value-theorem-1d-2026c
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Reason: Publish successor MVT proof using Rolle 2026c.

Proof

Define

m=f(b)βˆ’f(a)bβˆ’am=\frac{f(b)-f(a)}{b-a}

and set Ο•:Iβ†’R\phi:I\to\mathbb{R} by Ο•(x)=f(x)βˆ’mx\phi(x)=f(x)-mx. Since ff is continuous on [a,b][a,b] and differentiable at every point of (a,b)(a,b) by hypothesis, the same is true for Ο•\phi. Also,

Ο•(b)βˆ’Ο•(a)=(f(b)βˆ’mb)βˆ’(f(a)βˆ’ma)=f(b)βˆ’f(a)βˆ’m(bβˆ’a)=0,\phi(b)-\phi(a)=\bigl(f(b)-mb\bigr)-\bigl(f(a)-ma\bigr)=f(b)-f(a)-m(b-a)=0,

so Ο•(a)=Ο•(b)\phi(a)=\phi(b). Therefore Rolle's Theorem in One Dimension applies to Ο•\phi on [a,b][a,b], and there exists c∈(a,b)c\in(a,b) such that Ο•β€²(c)=0\phi'(c)=0. Because

Ο•β€²(c)=fβ€²(c)βˆ’m,\phi'(c)=f'(c)-m,

we obtain

fβ€²(c)=m=f(b)βˆ’f(a)bβˆ’a.f'(c)=m=\frac{f(b)-f(a)}{b-a}.

This is the desired conclusion.

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