6.3: Mean Value Theorem

• Dan Sloughter
• Furman University

Definition

We say \(f\) is differentiable on an open interval \(I\) if \(f\) is differentiable at every point \(a \in I\).

Definition

Suppose \(D \subset \mathbb\) and \(f: D \rightarrow \mathbb .\) We say \(f\) has a local maximum at a point \(a \in D\) if there exists \(\delta>0\) such that \(f(a) \geq f(x)\) for all \(x \in(a-\delta, a+\delta) \cap D .\) We say \(f\) has a local minimum at a point \(a \in D\) if there exists \(\delta>0\) such that \(f(a) \leq f(x)\) for all \(x \in(a-\delta, a+\delta) \cap D .\)

Proposition \(\PageIndex\)

Suppose \(D \subset \mathbb, f: D \rightarrow \mathbb,\) and \(a\) is an interior point of \(D\) at which \(f\) has either a local maximum or a local minimum. If \(f\) is differentiable at \(a,\) then \(f^<\prime>(a)=0\). Proof Suppose \(f\) has a local maximum at \(a\) (a similar argument works if \(f\) has a local minimum at \(a\) ). Choose \(\delta>0\) so that \((a-\delta, a+\delta) \subset D\) and \(f(a) \geq f(x)\) for all \(x \in(a-\delta, a+\delta) .\) Then \[\frac \geq 0\] for all \(x \in(a-\delta, a)\) and \[\frac \leq 0\] for all \(x \in(a, a+\delta) .\) Hence \[\lim _> \frac \geq 0\] and \[\lim _> \frac \leq 0.\] Hence \[0 \leq \lim _> \frac=f^<\prime>(a)=\lim _> \frac \leq 0,\] so we must have \(f^<\prime>(a)=0\). \(\quad\) Q.E.D.

Theorem \(\PageIndex\)

(Rolle's Theorem). Let \(a, b \in \mathbb\) and suppose \(f\) is continuous on \([a, b]\) and differentiable on \((a, b) .\) If \(f(a)=f(b),\) then there exists a point \(c \in(a, b)\) at which \(f^<\prime>(c)=0\). Proof By the Extreme Value Theorem, we know \(f\) attains a maximum and a minimum value on \([a, b] .\) Let \(m\) be the minimum value and \(M\) the maximum value of \(f\) on \([a, b] .\) If \(m=M=f(a)=f(b),\) then \(f(x)=m\) for all \(x \in[a, b],\) and so \(f^<\prime>(x)=0\) for all \(x \in(a, b) .\) Otherwise, one of \(m\) or \(M\) occurs at a point \(c\) in \((a, b) .\) Hence \(f\) has either a local maximum or a local minimum at \(c,\) and so \(f^<\prime>(c)=0 .\) \(\quad\) Q.E.D.

Exercise \(\PageIndex\)

Suppose \(f\) is differentiable on \((a, b)\) and \(f^<\prime>(x) \neq 0\) for all \(x \in(a, b) .\) Show that for any \(x, y \in(a, b), f(x) \neq f(y)\).

Exercise \(\PageIndex\)

Exercise \(\PageIndex\)

Let \(f(x)\) be a third degree polynomial. Show that the equation \(f(x)=0\) as at least one, but no more than three, solutions.

6.3.2 Mean Value Theorem

Theorem \(\PageIndex\)

(Generalized Mean Value Theorem). Let \(a, b \in \mathbb .\) If \(f\) and \(g\) are continuous on \([a, b]\) and differentiable on \((a, b),\) then there exists a point \(c \in(a, b)\) at which \[(f(b)-f(a)) g^<\prime>(c)=(g(b)-g(a)) f^<\prime>(c).\] Proof Let \[h(t)=(f(b)-f(a)) g(t)-(g(b)-g(a)) f(t).\] Then \(h\) is continuous on \([a, b]\) and differentiable on \((a, b) .\) Moreover, \[\begin h(a) &=f(b) g(a)-f(a) g(a)-f(a) g(b)+f(a) g(a) \\ &=f(b) g(a)-f(a) g(b) \end\] and \[\begin h(b) &=f(b) g(b)-f(a) g(b)-f(b) g(b)+f(b) g(a) \\ &=f(b) g(a)-f(a) g(b). \end\] Hence, by Rolle's theorem, there exists a point \(c \in(a, b)\) at which \(h^<\prime>(c)=0 .\) But then \[0=h^<\prime>(c)=(f(b)-f(a)) g^<\prime>(c)-(g(b)-g(a)) f^<\prime>(c),\] which implies that \[(f(b)-f(a)) g^<\prime>(c)=(g(b)-g(a)) f^<\prime>(c).\] Q.E.D.

Theorem \(\PageIndex\)

(Mean Value Theorem). Let \(a, b \in \mathbb .\) If \(f\) is continuous on \([a, b]\) and differentiable on \((a, b),\) then there exists a point \(c \in(a, b)\) at which \[f(b)-f(a)=(b-a) f^<\prime>(c).\] Proof Apply the previous result with \(g(x)=x\). \(\quad\) Q.E.D.

Exercise \(\PageIndex\)

Prove the Mean Value Theorem using Rolle's theorem and the function \[k(t)=f(t)-\left(\left(\frac\right)(t-a)+f(a)\right).\] Give a geometric interpretation for \(k\) and compare it with the function \(h\) used in the proof of the generalized mean value theorem.

Exercise \(\PageIndex\)

Let \(a, b \in \mathbb .\) Suppose \(f\) is continuous on \([a, b],\) differentiable on \((a, b),\) and \(\left|f^<\prime>(x)\right| \leq M\) for all \(x \in(a, b) .\) Show that \[|f(b)-f(a)| \leq M|b-a|.\]

Exercise \(\PageIndex\)

Exercise \(\PageIndex\)

Suppose \(I\) is an open interval, \(f: I \rightarrow \mathbb,\) and \(f^<\prime>(x)=0\) for all \(x \in I .\) Show that there exists \(\alpha \in \mathbb\) such that \(f(x)=\alpha\) for all \(x \in I\).

Exercise \(\PageIndex\)

Suppose \(I\) is an open interval, \(f: I \rightarrow \mathbb, g: I \rightarrow \mathbb,\) and \(f^<\prime>(x)=g^<\prime>(x)\) for all \(x \in I .\) Show that there exists \(\alpha \in \mathbb\) such that \[g(x)=f(x)+\alpha\] for all \(x \in I\).

Exercise \(\PageIndex\)

Let \(D=\mathbb \backslash\ .\) Define \(f: D \rightarrow \mathbb\) and \(g: D \rightarrow \mathbb\) by \(f(x)=x^\) and \[g(x)=\left\ & x \\ & x>0.>\end\right.\] Show that \(f^<\prime>(x)=g^<\prime>(x)\) for all \(x \in D,\) but there does not exist \(\alpha \in \mathbb\) such that \(g(x)=f(x)+\alpha\) for all \(x \in D .\) Why does this not contradict the conclusion of the previous exercise?

Proposition \(\PageIndex\)

If \(f\) is differentiable on \((a, b)\) and \(f^<\prime>(x)>0\) for all \(x \in(a, b)\), then \(f\) is increasing on \((a, b)\). Proof Let \(x, y \in(a, b)\) with \(x(c).\] Since \(y-x>0\) and \(f^<\prime>(c)>0,\) we have \(f(y)>f(x),\) and so \(f\) is increasing on \((a, b) .\) \(\quad\) Q.E.D.

Theorem \(\PageIndex\)

If \(f\) is differentiable on \((a, b)\) and \(f^<\prime>(x)

Exercise \(\PageIndex\)

This page titled 6.3: Mean Value Theorem is shared under a CC BY-NC-SA 1.0 license and was authored, remixed, and/or curated by Dan Sloughter via source content that was edited to the style and standards of the LibreTexts platform.

1. Back to top
2. • 6.2: Derivatives
   • 6.4: Discontinuities of Derivatives

• Was this article helpful?
• Yes
• No

Recommended articles

1. Article type Section or Page Author Dan Sloughter License CC BY-NC-SA License Version 1.0 Show Page TOC no
2. Tags
   1. source@http://www.synechism.org/wp/the-calculus-of-functions-of-several-variables