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Mathematics LibreTexts

6.3: Mean Value Theorem

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    22673
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    Definition

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

    Definition

    Suppose \(D \subset \mathbb{R}\) and \(f: D \rightarrow \mathbb{R} .\) 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 \(\PageIndex{1}\)

    Suppose \(D \subset \mathbb{R}, f: D \rightarrow \mathbb{R},\) 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

    \[\frac{f(x)-f(a)}{x-a} \geq 0\]

    for all \(x \in(a-\delta, a)\) and

    \[\frac{f(x)-f(a)}{x-a} \leq 0\]

    for all \(x \in(a, a+\delta) .\) Hence

    \[\lim _{x \rightarrow a^{-}} \frac{f(x)-f(a)}{x-a} \geq 0\]

    and

    \[\lim _{x \rightarrow a^{+}} \frac{f(x)-f(a)}{x-a} \leq 0.\]

    Hence

    \[0 \leq \lim _{x \rightarrow a^{-}} \frac{f(x)-f(a)}{x-a}=f^{\prime}(a)=\lim _{x \rightarrow a^{+}} \frac{f(x)-f(a)}{x-a} \leq 0,\]

    so we must have \(f^{\prime}(a)=0\). \(\quad\) Q.E.D.

    Theorem \(\PageIndex{2}\)

    (Rolle's Theorem).

    Let \(a, b \in \mathbb{R}\) 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 \(\PageIndex{1}\)

    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 \(\PageIndex{2}\)

    Explain why the equation \(x^{5}+10 x=5\) has exactly one solution.

    Exercise \(\PageIndex{3}\)

    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 \(\PageIndex{3}\)

    (Generalized Mean Value Theorem).

    Let \(a, b \in \mathbb{R} .\) 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,

    \[\begin{aligned} h(a) &=f(b) g(a)-f(a) g(a)-f(a) g(b)+f(a) g(a) \\ &=f(b) g(a)-f(a) g(b) \end{aligned}\]

    and

    \[\begin{aligned} h(b) &=f(b) g(b)-f(a) g(b)-f(b) g(b)+f(b) g(a) \\ &=f(b) g(a)-f(a) g(b). \end{aligned}\]

    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 \(\PageIndex{4}\)

    (Mean Value Theorem).

    Let \(a, b \in \mathbb{R} .\) 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 \(\PageIndex{4}\)

    Prove the Mean Value Theorem using Rolle's theorem and the function

    \[k(t)=f(t)-\left(\left(\frac{f(b)-f(a)}{b-a}\right)(t-a)+f(a)\right).\]

    Give a geometric interpretation for \(k\) and compare it with the function \(h\) used in the proof of the generalized mean value theorem.

    Exercise \(\PageIndex{5}\)

    Let \(a, b \in \mathbb{R} .\) 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 \(\PageIndex{6}\)

    Show that for all \(x>0\),

    \[\sqrt{1+x}<1+\frac{x}{2}.\]

    Exercise \(\PageIndex{7}\)

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

    Exercise \(\PageIndex{8}\)

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

    \[g(x)=f(x)+\alpha\]

    for all \(x \in I\).

    Exercise \(\PageIndex{9}\)

    Let \(D=\mathbb{R} \backslash\{0\} .\) Define \(f: D \rightarrow \mathbb{R}\) and \(g: D \rightarrow \mathbb{R}\) by \(f(x)=x^{2}\) and

    \[g(x)=\left\{\begin{array}{ll}{x^{2},} & {\text { if } x<0,} \\ {x^{2}+1,} & {\text { if } x>0.}\end{array}\right.\]

    Show that \(f^{\prime}(x)=g^{\prime}(x)\) for all \(x \in D,\) but there does not exist \(\alpha \in \mathbb{R}\) such that \(g(x)=f(x)+\alpha\) for all \(x \in D .\) Why does this not contradict the conclusion of the previous exercise?

    Proposition \(\PageIndex{5}\)

    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 \(x<y .\) By the Mean Value Theorem, there exists a point \(c \in(x, y)\) such that

    \[f(y)-f(x)=(y-x) f^{\prime}(c).\]

    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 \(\PageIndex{6}\)

    If \(f\) is differentiable on \((a, b)\) and \(f^{\prime}(x)<0\) for all \(x \in(a, b)\), then \(f\) is decreasing on \((a, b)\).

    Exercise \(\PageIndex{10}\)

    State and prove similar conditions for nonincreasing and nondecreasing functions.

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