Mean Value Theorem

This section demonstrates some examples that do not satisfy some of the conditions of our calculus theorems.

1. Discontinuity. Example: $f(x)=a(x-b)^2+c$ except for $x=b$
2. Non-differentiability. Example:
\begin{align} f(x) & = \left\{ \begin{array}{ll} -a_1 |\sin(\frac{\pi x}{6})|+b_1 & x \lt 6; \\ -a_2 |\sin(\frac{\pi x}{6})|+b_2 & x \gt 6; \\ \end{array}\right. \end{align}
3. Unequal function values at end points of an interval. Example: $f(x)=e^{\frac{x}{10}}$

Examples.

1. Let $f(x)=x^3-x^2-x+k$. Prove that there is a root of $f^{\prime}(x)$ in $(-1,1)$.

Solution.

$f^{\prime}(x) =$$3x^2-2x-1$$x^2-x-1$ $3x^2-2x-1 = $$(3x-1)(x+1)$$(3x+1)(x-1)$$(3x+1)(x-1) = 0$Oops! Try again

$x = 1 \notin (-1,1) \text{ or } x = -\frac{1}{3} \in (-1,1)$

Thus, the derivative of the function has a root in $(-1,1)$.





2. Using Lagrange's mean value theorem, show that $\frac{b-a}{b} \lt \ln (\frac{b}{a} ) \lt \frac{b-a}{a}$ where $0 \lt a \lt b.$

   Solution.

   Let $f(x) = \ln x, 0\lt a \lt b$.

   Clearly, $f(x)$ is continuous in $[a,b]$ and differentiable in $(a,b)$, with $f^{\prime}(x)=\frac{1}{x}$.

   Next

   By Lagrange's mean value theorem, $\frac{1}{c} = \frac{ \ln (\frac{b}{a})}{b-a}$

   Next

   Now, since $a\lt c \lt b$,

   $\frac{1}{a} \lt \frac{1}{c} \lt \frac{1}{b}$$\frac{1}{b} \lt \frac{1}{c} \lt \frac{1}{a}$Oops! Try again

   \begin{align} \frac{1}{b} & \lt \frac{1}{c} \lt \frac{1}{a} \\ \frac{1}{b} & \lt \frac{\ln (\frac{b}{a} )}{b-a} \lt \frac{1}{a} \\ \frac{b-a}{b} & \lt \ln ( \frac{b}{a} ) \lt \frac{b-a}{a} \end{align}

   
   
   
   
3. Using Cauchy's mean value theorem, show that if $0 \lt \alpha \lt \beta \lt \frac{\pi}{2}$, then there exists some $\theta \in (\alpha, \beta)$ such that $\frac{\sin \beta - \sin \alpha}{\cos \beta - \cos \alpha} = -\cot \theta$.

Solution.

Let $f(x)=\sin x, g(x)=\cos x$.

Check all the conditions for Cauchy's MVT

Clearly, $f(x)$ and $g(x)$ are continuous in $[\alpha,\beta]$ and differentiable in $(\alpha,\beta)$.

Also, $g'(x)=-\sin x \neq 0$ for any $x \in (0,\frac{\pi}{2})$

Next

By Cauchy's mean value theorem,

\begin{align} \frac{f'(\theta)}{g'(\theta)} & = \frac{f(\beta)-f(\alpha)}{g(\beta)-g(\alpha)}, 0 \lt \alpha \lt \theta \lt \beta \lt \frac{\pi}{2} \\ \frac{\sin \beta - \sin \alpha}{\cos \beta - \cos \alpha} & = \frac{\cos \theta}{-\sin \theta} = \cot \theta \end{align}

Examples.

1. Verify Rolle's Theorem for the function $f(x)=\ln ( \frac{x^2+ab}{x(a+b)} )$ in $[a,b]$, where $0\lt a\lt b.$

Solution.

Conditions:

Conditions:

The logarithmic function is always continuous and differentiable in its domain.

Therefore, $f(x)$ is continuous in $[a,b]$ and differentiable in $(a,b)$.

Also, $f(a)=\ln ( \frac{a^2+ab}{a(a+b)} ) = \ln 1 = 0$ and $f(b) = \ln (\frac{b^2+ab}{b(a+b)} ) = \ln 1 = 0$.

Conclusion:

Conclusion:

Need to show: that there exists a point $c \in (a,b)$ such that $f'(c)=0$.

\begin{align} f^{\prime}(x) & = \frac{2x}{x^2+ab}-\frac{1}{x} \\ & = \frac{2x^2-x^2-ab}{x(x^2+ab)} \\ & = \frac{x^2-ab}{x(x^2)+ab} \\ f^{\prime}(c) & = 0 \\ \Rightarrow c & = \sqrt{ab} \in (a,b). \end{align}




2. Verify Lagrange's mean value theorem for the function $f(x)=x^3-x^2-x+k$ on $[0,2]$.

Solution.

Conditions:

Conditions:

Since $f(x)$ is a polynomial, it is continuous in $[0,2]$ and differentiable in $(0,2)$.

Conclusion:

Conclusion:

Need to show: that there exists a point $c \in (a,b)$ such that $f^{\prime}(c)=\frac{f(2)-f(0)}{2-0}=\frac{3-1}{2}=1$.

\begin{align} f^{\prime}(c) = 3c^2-2c-1 & = 1 \\ 3c^2-2c-2 & = 0 \\ c & = \frac{1 \pm \sqrt{7}}{3} \in (0,2) \end{align}




3. Verify Cauchy's mean value theorem for the functions $f(x)=kx^2$ and $g(x)=kx^3$ in the interval $[1,2]$.

Solution.

Conditions:

Conditions:

Since $f(x)$ and $g(x)$ are polynomials, they are continuous in $[1,2]$ and differentiable in $(1,2)$.

Also, $g'(x)=3kx^2\neq0$ for any $x\in(1,2)$.

Conclusion:

Conclusion:

Need to show: that there exists a point $c \in (a,b)$ such that $\frac{f'(c)}{g'(c)}=\frac{f(2)-f(1)}{g(2)-g(1)}$.

\begin{align} \frac{4-1}{8-1} & = \frac{2c}{3c^2} \\ 9c^2-14c & = 0 \\ c & = 0 \text{ or } c = \frac{14}{9} \in (1,2) \end{align}

Examples.

1. The nature of the cubic polynomial $f(x)=ax^3+bx^2+cx+d$.

The coefficients $a,b,c,d$ determine the number of $\textbf{real}$ roots of $f(x)$.

(Remark: By the fundamental theorem of algebra, any polynomial of degree $n$ has $n$ roots in $\textbf{complex}$ numbers)

We may analyze this using differentiation and our understanding of quadratic equations: $f'(x)=3ax^2+2bx+c$:

1. $f'(x)$ has no real root. Then $f(x)$ is always increasing and the equation $f(x)=0$ has one real root.
2. $f'(x)$ has two distinct real roots $x_1$, $x_2$ and $f(x_1)f(x_2) \gt 0$. Then the equation $f(x)=0$ has one real root.
3. $f'(x)$ has two distinct real roots $x_1$, $x_2$ and $f(x_1)f(x_2) \lt 0$. Then the equation $f(x)=0$ has three distinct real roots.
4. $f'(x)$ has two distinct real roots $x_1$, $x_2$ and $f(x_1)f(x_2) = 0$. Then the equation $f(x)=0$ has two distinct real roots.
5. $f'(x)$ has two equal real roots $\alpha$ and $f(\alpha)=0.$ Then the equation has one real repeated root.




2. By considering the function $f(x)=e^x$, show that $$1+x < e^x < 1+xe^x \quad \forall x>0.$$

Solution.

Left inequalityRight inequality

The exponential function is defined by $e^x=1+x+\frac{x^2}{2!}+\cdots > 1+x$ for any $x>0$.

For any real number $x>0$, $f(x)=e^x$ is continuous in $[0,x]$ and differentiable in $(0,x)$.

By Lagrange's mean value theorem, there exists $c \in (0,x)$ such that

\begin{align} f(x)-f(0) & =(x-0)f^{\prime}(0+c) \\ e^x-e^0 & =xf^{\prime}(c) \\ e^x & =1+xe^c < 1+xe^x \quad \text{since } c < x \end{align}




3. Evaluate $$\lim \limits_{x\rightarrow 0} \frac{\cos \frac{1}{2}\pi x}{\ln (\frac{1}{x})}.$$

Solution.

Let $f(x)=\cos(\frac{1}{2}\pi x)$ and $\phi (x)=\ln x$.

Applying Cauchy's mean value theorem in $[x,1]$ with $x \in (0,1)$, we have

\begin{align} \frac{\cos(\frac{\pi}{2})-\cos(\frac{\pi}{2}x)}{\ln 1-\ln x} & = \frac{-\frac{1}{2}\pi\sin(\frac{\pi}{2}c)}{1/c} \quad \text{for some } x < c < 1 \\ \frac{\cos(\frac{1}{2}\pi x)}{\ln(\frac{1}{x})} & =\frac{\pi c}{2}\sin(\frac{\pi c}{2}) \end{align} Next

If $x\to 1$, then $c\to 1$ and we have

$$\lim\limits_{x\rightarrow 1}\frac{\cos\frac{\pi}{2}x}{\ln\frac{1}{x}}=\lim\limits_{c\rightarrow 1}\frac{\pi c}{2}\sin\left(\frac{\pi c}{2}\right) = \lim\limits_{c\rightarrow 1}\frac{\pi c}{2} \cdot \lim\limits_{c\rightarrow 1}\sin\left(\frac{\pi c}{2}\right) = \frac{\pi}{2}.$$

Remark.

This example demonstrates the relationship between Cauchy's mean value theorem and L'Hopital's rule.

Exercises.

1. By considering the function $f(x)=(x-4)\ln x$, prove that the equation $$x\ln x=4-x$$ is satisfied by at least one value of $c\in (1,4)$.

Solution.

1. Since $f$ is the product of two continuous functions in $[1,4]$, $f$ is continuous in $[1,4]$.
2. The derivative of $f(x)$ is $f^{\prime}(x)=\dfrac{x-4}{x}+\ln x$ (using the product rule), which exists finitely in $[1,4].$ Therefore, $f(x)$ is differentiable in $(1,4)$.
3. We have $f(1) = (1-4)\ln 1=0, f(4)=(4-4)\ln 4=0.$ Hence, $f(1)=f(4)$.

Hence by Rolle's theorem Lagrage's mean value theorem Cauchy's mean value theorem continuous differentiable constant bounded , there exists some $c\in (1,4)$ such that $f^{\prime}(c)=\frac{c-4}{4}+\ln c=0$ which gives $4\ln c=4-c$.

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Remark.

One may location the root by using numerical methods like the Newton-Raphson method and the bisection method.





2. Let $0\lt a \lt b$. Show that $\frac{b-a}{1+b^2} \lt \tan^{-1} b-\tan^{-1} a \lt \frac{b-a}{1+a^2}$.

Solution.

Let $f(x)=\tan^{-1} x, x\in [a,b].$ For any $0 \lt a \lt b, f(x)$ is (i) Rolle's theorem Lagrage's mean value theorem Cauchy's mean value theorem continuous differentiable constant bounded in $[a,b]$ and (ii) Rolle's theorem Lagrage's mean value theorem Cauchy's mean value theorem continuous differentiable constant bounded in $(a,b)$. Then by (iii) Rolle's theorem Lagrage's mean value theorem Cauchy's mean value theorem continuous differentiable constant bounded , there exists $a \lt c \lt b$ such that $f'(c)=\frac{\tan^{-1} b-\tan^{-1} a}{b-a}$.

Also, $f'(x)=\frac{1}{1+x^2}$. Since $f'(x)$ is monotonically decreasing, $a \lt c \lt b \Rightarrow f'(b) \lt f'(c) \lt f'(a)$

Therefore, $\frac{b-a}{1+b^2} \lt \tan^{-1} b-\tan^{-1} a \lt \frac{b-a}{1+a^2}$.

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3. Let $a,b$ be real numbers and $a \lt b$. Determine the values of $c$ satisfying Cauchy's mean value theorem for the given pairs of functions in $[a,b]$ and match them with the following values.

(a) $\frac{a+b}{2}$ (b) $\sqrt{ab}$ (c) $\frac{2ab}{a+b}$

1. $f(x)=\sqrt{x}, g(x)=\frac{1}{\sqrt{x}}$ (a) (b) (c)
2. $f(x)=\frac{1}{x^2}, g(x)=\frac{1}{x}$ (a) (b) (c)
3. $f(x)=e^x, g(x)=e^{-x}$ (a) (b) (c)

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Show solution

Recall Cauchy's mean value theorem: $\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(c)}{g'(c)}$
1. (b). \begin{align} \frac{\frac{1}{b^2}-\frac{1}{a^2}}{\frac{1}{b}-\frac{1}{a}} & = \frac{-2c^{-3}}{-c^{-2}} \\ \frac{1}{ab} \frac{a^2-b^2}{a-b} & = \frac{2}{c} \\ c & = \frac{2ab}{a+b} \\ \end{align}
2. (c). \begin{align} \frac{\sqrt{b}-\sqrt{a}}{\frac{1}{\sqrt{b}}-\frac{1}{\sqrt{a}}} & = \frac{\frac{1}{2}c^{-1/2}}{-\frac{1}{2}c^{-3/2}} \\ c & = \sqrt{ab} \end{align}
3. (a). \begin{align} \frac{e^b-e^a}{e^{-b}-e^{-a}} & = \frac{e^c}{e^{-c}} \\ -e^{a+b} & = -e^{2c} \\ c & = \frac{a+b}{2} \\ \end{align}

Remark.

The quantities (a), (b), (c) are respectively the arithmetic mean, geometric mean and harmonic mean of $a$ and $b$. In general,

• The arithmetic mean of $x_1, \cdots , x_n$ is defined by $\frac{1}{n}(x_1+ \cdots +x_n)$.
• The geometric mean of $x_1, \cdots, x_n$ is defined by $(x_1 \cdots x_n)^{\frac{1}{n}}$.
• The harmonic mean of $x_1, \cdots, x_n$ is defined by $N/(\frac{1}{x_1}+ \cdots +\frac{1}{x_n})$.

Exercises.

1. To which of the following functions is/are Rolle's theorem applicable on the given interval?
1. $f(x) = \left\{ \begin{array}{ll} x & 0 \leq x \lt 1 \\ 0 & x=1 \\ \end{array}\right. \text{ on } [0,1]$
2. $g(x) = \left\{ \begin{array}{ll} \frac{\sin x}{x} & -\pi \leq x \lt 0 \\ 0 & x=0 \\ \end{array}\right. \text{ on } [-\pi,0]$
3. $h(x) = \frac{x^2-x-6}{x-1} \text{ on } [-2,3]$
4. $k(x) = \left\{ \begin{array}{ll} \frac{x^3-2x^2-5x+6}{-6} & x\neq 1 \\ -6 & x=1 \\ \end{array}\right. \text{ on } [-2,3]$

Check answers

See explanation

1. Not applicable. $f(x)$ is not continuous at $x=1$.
2. Not applicable. $g(x)$ is not continuous at $x=0$.
3. Not applicable. $h(x)$ is not continuous at $x=1$.
4. Applicable.




2. To which of the following is/are Lagrange's mean value theorem NOT applicable on $[0,1]$?
1. $f(x) = 4-(\frac{1}{2}-x)^{\frac{2}{3}}$
2. $g(x) = \left\{ \begin{array}{ll} \frac{\tan x}{x} & x \neq 0 \\ 1 & x=0 \\ \end{array}\right.$
3. $h(x) = 5^{\log_2 (x+3)}$
4. $k(x) = \left\{ \begin{array}{ll} \frac{1}{2}-x & x\lt \frac{1}{2} \\ (\frac{1}{2}-x)^2 & x\leq \frac{1}{2} \\ \end{array}\right.$

Check answers

See explanation

1. Not applicable. $f(x)$ is not differentiable at $x=\frac{1}{2}$.
2. Applicable.
3. Applicable.
4. Not applicable. $k(x)$ is not differentiable at $x=\frac{1}{2}$.




3. Consider $f(x)=|1-x|, \quad g(x)=f(x)+b \sin(\frac{\pi x}{2})$, where $b\in \mathbb{R}$. Which of the following statements is/are correct for the interval $[1,2]$?
1. Rolle's theorem is applicable to both $f(x)$ and $g(x)$ with $b=\frac{3}{2}$.
2. Lagrange's mean value theorem is not applicable to $f(x)$ but Rolle's theorem is applicable to $g(x)$ with $b=\frac{1}{2}$.
3. Lagrange's mean value theorem is applicable to $f(x)$ and Rolle's theorem is applicable to $g(x)$ with $b=1$.
4. Rolle's theorem is not applicable to either $f(x)$ or $g(x)$ for any $b\in \mathbb{R}$.

Check answers

See explanation

$f(x)=x-1, g(x)=x-1+b \sin(\frac{\pi x}{2}), 1\leq x \leq 2$.
The functions $x-1$ and $\sin(\frac{\pi x}{2})$ are continuous in $[1,2]$ and $x-1$ is differentiable in $(1,2)$.
$f(1)=0, f(2)=1 \Rightarrow $ Rolle's theorem is not applicable to $f(x)$ but Lagrange's mean value theorem is applicable to $f(x)$.
$g(1)=b, g(2)=1 \Rightarrow $ For Rolle's theorem to be applicable to $g(x)$, we must have $b=1$.
Hence, only option C is correct.

Examples.

1. Rolle's Theorem: continuity + differentiability + equal end point function values $\Rightarrow$ existence of horizontal tangent.

Example: $f(x)=a(x-b)^2+c$

2. Lagrange's Mean Value Theorem: continuity + differentiability $\Rightarrow$ existence of some tangent line which is parallel to secant line passing through the end points of an interval.

Example: $f(x)=e^{\frac{x}{10}}$

3. Cauchy's Mean Value Theorem: continuity + differentiability $\Rightarrow$ There is a point $c$ such that the ratio of the derivatives of $f$ and $g$ is equal to the ratio of the differences of $f$ and $g$ in an interval

Example (i): $f(x)=k(-x^2+8), g(x)=k(2x^2-4)$

$f(0)=4k, f(2)=8k, g(0)=-4k, g(2)=-4k$

$g(2)-g(0)=2(f(2)-f(0))$

Example (ii): $f(x)=k(\frac{x^3}{6}+2), g(x)=k(\frac{(x+2)^3}{3}-2)$

$g(2)-g(0)=4(f(2)-f(0))$

Exercises.

1. Prove that there does not exist any real $\lambda$ for which the equation $x^3-27x+\lambda =0$ has two different roots in $[0,3].$

Solution.

We will prove the desired reult by contradiction. Suppose instead that for some $\lambda \in \mathbb{R}$, the equation $x^3-27x+\lambda=0$ has two distinct roots $\alpha \beta\in [0,3]$, where $0 \leq \alpha < \beta \leq 3$.

Now, consider a function $f(x)$ in closed interval $[\alpha ,\beta ]$:

$$f(x)=x^3-27x+\lambda\ \ \ \ \ \ \ \ \forall x\in [\alpha ,\beta].$$
1. Being a polynomial, $f(x)$ is continuous in $[\alpha ,\beta ]$.
2. $f^{\prime}(x)=3x^2-27$ exists finitely in $(\alpha ,\beta ).$ Therefore, $f(x)$ is differentiable in $(\alpha ,\beta ).$
3. Since $\alpha$, $\beta$ are roots of $f(x), f(\alpha)=f(\beta)=0$.

Hence by Rolle's theorem Lagrage's mean value theorem Cauchy's mean value theorem continuous differentiable constant bounded , there exists some $c\in (\alpha,\beta)$ such that $f^{\prime}(c)=3c^2-27=0$

Then $c=\pm 3 \notin (0,3)$ implies that $c\notin (\alpha,\beta)$, which is a contradiction.

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2. If $f$ is a function such that $f^{\prime}(x)$ exists in $[a,b]$ and $$\frac{f(c)-f(a)}{c-a}=\frac{f(b)-f(c)}{b-c},$$ where $c\in (a,b)$, show that there exists $\alpha\in (a,b)$ such that $f^{\prime}(\alpha )=0$.

Solution.

It is given that $f^{\prime}$ exists in $[a,b]$. Therefore, $f$ and $f^{\prime}$ are both continuous in $[a,b]$ since $c\in (a,b).$

Applying (i) Rolle's theorem Lagrage's mean value theorem Cauchy's mean value theorem continuous differentiable constant bounded in $[a,c]$ and $[c,b]$ respectively, we have

\begin{align} \dfrac{f(c)-f(a)}{c-a} & =f^{\prime}(\alpha_1) && \text{for some } a < \alpha_1 < c \\ \dfrac{f(b)-f(c)}{b-c} & =f^{\prime}(\alpha_2) && \text{for some } c < \alpha_2 < b \end{align}

which means that $[\alpha_1,\alpha_2] \subset [a,b]$. Now, since $\frac{f(c)-f(a)}{c-a}=\frac{f(b)-f(c)}{b-c}$, we must have $f^{\prime}(\alpha_1)=f^{\prime}(\alpha_2).$

Also, $f^{\prime}(x)$ is continuous in $[\alpha_1,\alpha_2]$ and $f^{\prime}(x)$ exists in $(\alpha_1,\alpha_2)$ means that $f^{\prime}(x)$ is differentiable.

Thus by (ii) Rolle's theorem Lagrage's mean value theorem Cauchy's mean value theorem continuous differentiable constant bounded , there exists $\alpha\in (\alpha_1,\alpha_2)\subset [a,b]$ such that $f^{\prime}(\alpha )=0$.

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3. Let $f$ be a continuous and differentiable function in $[a,b]$. Show that for some $c\in (a,b)$, $$f(b)-f(a)=cf^{\prime}(c)\ln\frac{b}{a}.$$

Solution.

Let $g(x)=\ln x$. Then $g^{\prime}(c)=\frac{1}{c}$.

By Rolle's theorem Lagrage's mean value theorem Cauchy's mean value theorem continuous differentiable constant bounded , we have

\begin{align} \frac{f(b)-f(a)}{\ln b-\ln a} & =\frac{f^{\prime}(c)}{1/c} \\ \frac{f(b)-f(a)}{\ln\frac{b}{a}} & = cf^{\prime}(c) \\ f(b)-f(a) & =cf^{\prime}(c)\ln\frac{b}{a}. \end{align}

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Definitions.

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a function.

1. (Limit) A real number $L$ is said to be a limit of $f$ at $c$ if \begin{align} \text{For any } \epsilon >0, \text{ there exists } \delta >0 \text{ such that if } |x-c| \lt \delta, \text{ then } |f(x)-L|\lt \epsilon. \end{align}
2. (Continuity) $f$ is said to be continuous at $c$ if $\displaystyle \lim_{x\rightarrow c} f(x)=f(c)$.
3. (Differentiability) $f$ is said to be differentiable at $c$ if $\displaystyle f'(c)=\lim_{x\rightarrow c} \frac{f(x)-f(c)}{x-c}$ exists. $f'(c)$ is called the derivative of $f$ at $c$.

Theorem. (Rolle's Theorem)

If $f$ is a function defined on the closed interval $[a,b]$ such that

(i) $f$ is continuous in the closed interval $[a,b]$,
(ii) $f$ is differentiable in the open interval $(a,b)$, and
(iii) $f(a)=f(b)$,

then there exists some real number $c\in (a,b)$ such that $f^{\prime}(c)=0.$

Proof:

Since $f$ is (i) Rolle's theorem Lagrage's mean value theorem Cauchy's mean value theorem continuous differentiable constant bounded in $[a,b]$, $f$ is bounded and attains its bounds in the closed interval $[a,b]$.

That is, if $M$ and $m$ are the supremum and infimum of $f$ in $[a,b]$, then there exists $ c, d\in [a,b]$ such that

$$f(c)=m~~ {\rm and}~ ~ f(d)=M.$$

Case 1: $m=M$. Then $f$ is (ii) Rolle's theorem Lagrage's mean value theorem Cauchy's mean value theorem continuous differentiable constant bounded in $[a,b]$; this implies that $f^{\prime}(x)=0~\forall ~x\in [a,b].$

Case 2: $m\neq M.$ Let $m=f(c)\neq f(a)=f(b)$. Then $c\in (a,b)$. This implies that

$$f(x)\geq f(c), \quad \forall x\in [a,b].$$

Therefore, $f(c-h)\geq f(c)$ and $f(c+h)\geq f(c)$ for all $h>0$ such that $c\pm h\in [a,b]$. Hence, we have

$$\frac{f(c-h)-f(c)}{-h}\leq 0 \quad {\rm{and}} \quad \frac{f(c+h)-f(c)}{h}\geq 0.$$ $$\lim\limits_{h\rightarrow 0}\frac{f(c-h)-f(c)}{-h}\leq 0 \quad {\rm{and}} \quad \lim\limits_{h\rightarrow 0}\frac{f(c+h)-f(c)}{h}\geq 0$$

Since $f$ is (iii) Rolle's theorem Lagrage's mean value theorem Cauchy's mean value theorem continuous differentiable constant bounded at $x=c\in (a,b)$, we must have

$$f^{\prime}(c)=\lim\limits_{h\rightarrow 0}\frac{f(c-h)-f(c)}{-h}=\lim\limits_{h\rightarrow 0}\frac{f(c+h)-f(c)}{h} = 0$$

If $m=f(a)=f(b)$, then we may prove the theorem similarly by using $M=f(c)\neq f(a)=f(b)$.

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Theorem. (Lagrange's Mean Value Theorem)

If $f$ is a function defined on the closed interval $[a,b]$ such that

(i) $f$ is continuous in the closed interval $[a,b]$, and
(ii) $f$ is differentiable in the open interval $(a,b)$,

then there exists some real number $c\in (a,b)$ such that $\frac{f(b)-f(a)}{b-a}=f^{\prime}(c).$

Proof:

We want to define a new function $\psi(x)$ in the closed interval $[a,b]$ which involves the function $f(x)$ and satisfies all the conditions of Rolle's theorem.

Let $\psi (x)=f(x)+kx$, where $k$ is a constant. By construction, $\psi(x)$ is (i) Rolle's theorem Lagrage's mean value theorem Cauchy's mean value theorem continuous differentiable constant bounded in $[a,b]$ and (ii) Rolle's theorem Lagrage's mean value theorem Cauchy's mean value theorem continuous differentiable constant bounded in $(a,b)$.

In order that $\psi(x)$ satisfies $\psi(a)=\psi(b)$, we require $f(a)+ka = f(b)+kb$ which implies

$$k = -\frac{f(b)-f(a)}{b-a}, \quad a\neq b.$$

Then by (iii) Rolle's theorem Lagrage's mean value theorem Cauchy's mean value theorem continuous differentiable constant bounded , there exists $c\in (a,b)$ such that $\psi^{\prime}(c)=0.$ Now, we have

\begin{align} \psi^{\prime}(x) & = f^{\prime}(x)+k \\ \psi^{\prime}(c) & = f^{\prime}(c)+k = 0 \\ f^{\prime}(c) & = -k = \frac{f(b)-f(a)}{b-a} \end{align}

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Theorem. (Cauchy's Mean Value Theorem)

If $f$ and $g$ are two functions defined on the closed interval $[a,b]$ such that

(i) $f$ and $g$ are continuous on $[a,b]$,
(ii) $f$ and $g$ are differentiable on $(a,b)$, and
(iii) $g^{\prime}(x)\neq 0$ for $\forall x\in (a,b)$,

then there exists at least one point $c\in (a,b)$ such that $$\frac{f^{\prime}(c)}{g^{\prime}(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}.$$

Proof:

Firstly, we claim that $g(a)\neq g(b)$: If $g(a)=g(b)$, then by (i) Rolle's theorem Lagrage's mean value theorem Cauchy's mean value theorem continuous differentiable constant bounded , there exists $c\in (a,b)$ such that $g^{\prime}(c)=0$, which contradicts (iii).

Now let us define a new function $h(x)$ in the closed interval $[a,b]$ which involves $f$ and $g$ and satisfies all the conditions of Rolle's theorem.

Let $h(x)=f(x)+k\cdot g(x)$, where $k$ is a constant. By construction, $h(x)$ is (ii) Rolle's theorem Lagrage's mean value theorem Cauchy's mean value theorem continuous differentiable constant bounded in $[a,b]$ and (iii) Rolle's theorem Lagrage's mean value theorem Cauchy's mean value theorem continuous differentiable constant bounded in $(a,b)$.

In order that $h(x)$ satisfies $h(a)=h(b)$, we require $f(a)+k \cdot g(a) = f(b)+k \cdot h(b)$ which implies

$$k = -\frac{f(b)-f(a)}{g(b)-g(a)}.$$

Here, $k$ exists because $g(a)\neq g(a).$

Then by (iv) Rolle's theorem Lagrage's mean value theorem Cauchy's mean value theorem continuous differentiable constant bounded , there exists $c\in (a,b)$ such that $h^{\prime}(c)=0.$ Now, we have

\begin{align} h^{\prime}(x) & = f^{\prime}(x)+k \cdot g^{\prime}(x) \\ h^{\prime}(c) & = f^{\prime}(c)+k \cdot g^{\prime}(c)= 0 \\ \frac{f^{\prime}(c)}{g^{\prime}(c)} & = -k = \frac{f(b)-f(a)}{g(b)-g(a)} \end{align}

Check answers

Remark.

Lagrange's mean value theorem is a particular case of Cauchy's mean value theorem. Verify this by letting $g(x)=x$.

Remark.

You may have noticed that the formal definitions above were not used in our theorems.

You may learn rigorous proofs for these theorems in elementary analysis courses $\text{MATH}2050$ and $\text{MATH}2060$.

Aspiring MATH majors and anyone interested are encouraged to read the textbook "Introduction to Real Analysis".

Appendix. (General Mean Value Theorem)

If $f(x), g(x), h(x)$ be functions defined in the closed interval $[a,b]$ such that

(i) $f, g, h$ are all continuous in the closed interval $[a,b]$, and
(ii) $f, g, h$ are all differentiable in the open interval $(a,b)$,

then there exists $c \in (a,b)$ such that $$\left|\begin{array}{ccc} f^{\prime}(c)&g^{\prime}(c)&h^{\prime}(c)\\ f(a)&g(a)&h(a)\\ f(b)&g(b)&h(b) \end{array}\right|=0.$$

Proof.

Let us define a new function $\phi$ in the interval $[a,b]$ as follows: $$\phi (x)=\left|\begin{array}{ccc} f(x)&g(x)&h(x)\\ f(a)&g(a)&h(a)\\ f(b)&g(b)&h(b) \end{array}\right|.$$ Then $\phi$ is a linear function $$\phi (x)=\alpha f(x)+ \beta g(x)+ \gamma h(x).$$

The function $\phi$ is continuous in $[a,b]$ because it is a linear combination of continuous functions $f,g$ and $h$. Then, it is differentiable in $(a,b)$. Also, $\phi (a)=\phi (b)=0$. By Rolle's theorem, there exists at least one $c\in [a,b]$ such that $\phi^{\prime}(c)=0.$ That is, $$\left|\begin{array}{ccc} f^{\prime}(c)&g^{\prime}(c)&h^{\prime}(c)\\ f(a)&g(a)&h(a)\\ f(b)&g(b)&h(b) \end{array}\right|=0.$$

Remark.

Let $h(x)=k$ be constant and expand the determinant. What do you observe?