Roll's theorem, Mean Value Theorem, and the Shape of a Graph

Mean Value Theorem

From Bolzano-Weierstrass to Roll’s theorem

Theorem(Bolzano-Weierstrass) Let ${x_n}$ is a sequence in an closed interval $[a,b]$, then there exists a convergent subsequence of ${x_n}$.

proof. Let $a_0=a$ and $b_0=b$. We will recursively define $a_i$ and $b_i$. First, we consider two intervals $[a_0,\frac{a_0+b_0}{2}]$ and $[\frac{a_0+b_0}{2},b_0]$. Since ${x_n}$ has infinitely many terms, one of the intervals, denoted by $I=[a_1,b_1]$, contains infinitely many terms of the sequence. We then can find the smallest integer $n_1$ such that $x_{n_1}$ is in $I$. By iterating the above process, we will get three sequences ${a_i}$, ${b_i}$, and ${x_{n_i}}$, where ${x_{n_i}}$ is a subsequence of ${x_n}$. Note that $a_i\leq x_i\leq b_i$ and $\lim_{i\to\infty}a_i =\lim_{i\to\infty} b_i.$ By the squeeze theorem, $\displaystyle\lim_{i\to\infty}x_{n_i}$ exists.<div style="text-align: right;">∎</div>

Theorem(Weierstrass’s extreme value theorem) If $f(x)$ is a continuous function on a closed interval $[a,b]$, then $f(x)$ attains its local extreme values on $[a,b]$.

proof. For the sake of contradiction, we suppose we cannot find a maximum of $f$ on $[a,b]$. It is equivalent to say, for any positive integer $n$, we can find an $x_n\in[a,b]$ such that $f(x_n)>n$. By Bolzano-Weierstrass’s theorem, we have a convergent subsequence ${x_{n_i}}$ of ${x_n}$. Note that $f(x_{n_i})\geq n_i$, so $f(x_{n_i})\to\infty$ as $x_{n_i}\to x’$ for some $x’\in [a,b]$. This contradicts to the fact that $f(x)$ is continuous on $[a,b]$. Therefore, we must have a local maximum. Since $-f$ also satisfyies the assumption of this theorem, $-f$ has a local maximum in $[a,b]$, which implies $f$ has a local minimum in $[a,b]$.<div style="text-align: right;">∎</div>

Theorem(Fermat) If $f(x)$ is a continuous function on a closed interval with $f(c)$ an extreme values of $f$, and $f’(c)$ exists, then $f’(c)=0$.

Theorem(Fermat’s theorem in terms of critical points) If $f(x)$ is a continuous function on a closed interval with $f(c)$ an extreme values of $f$, then $f$ is a critical points. In particular, $\{x

f(x)\text{ is a local extreme}\}\subseteq\{x

x\text{ is a critical point of }f\}$.

proof. Weierstrass’s theorem implies that such a $c$ exists. Thus, we only need to show that $f’(c)=0$. We suppose it is a local maximum, i.e. $f(c)\geq f(x)$ for all $x$ near $c$. We then have $\frac{f(x)-f(c)}{x-c}\geq 0$ when $x<c$ and $x$ is near $c$. Thus, $\displaystyle\lim_{x\to c^-}\frac{f(x)-f(c)}{x-c}\geq 0$. On the other hand, we have $\frac{f(x)-f(c)}{x-c}\leq 0$ when $x>c$ and $x$ is near $c$. Thus, $\displaystyle\lim_{x\to c^+}\frac{f(x)-f(c)}{x-c}\leq 0$. Since we assume $f’(c)$ exists, the left limit has to be equal to the right limit, and we conclude that $f’(c)=0$.<div style="text-align: right;">∎</div>

Theorem(Roll) Suppose $f(x)$ satisfies the following three hypotheses:

$f(x)$ is a continuous function on $[a,b]$.
$f’(x)$ exists on $(a,b)$.
$f(a)=f(b)$. Then, there exists $c\in (a,b)$ such that $f’(c)=0$.

proof. Such a $c$ exists in $[a,b]$ by Fermat’s theorem. We only need to show that $c$ is not $a$ or $b$. Suppose $f$ is a constant function, then we have $f’(x)=0$ at any point in $(a,b)$. Otherwise, $f(a)$ and $f(b)$ cannot be a global maximum or minimum. Suppose it is not a globel maximum, then there exists $c\in(a,b)$ such that $f(c)$ is a local maximum and $f(c)>f(a)=f(b)$. Thus, we find a $c$ in the interial of $[a,b]$. Similar argument will work if we assume $f(a)$ and $f(b)$ is not a globle minimum.<div style="text-align: right;">∎</div>

Theorem(Mean value theorem) If $f(x)$ is continuous on $[a,b]$ and is differentiable on $(a,b)$, then there exists $c\in(a,b)$ such that $f'(c)=\frac{f(a)-f(b)}{a-b}$.

proof. Let $g(x)=f(x)-f(a)-\frac{f(b)-f(a)}{b-a}(x-a)$. We note that $g(a)=g(b)=0$. By Roll’s theorem, there exists $c\in(a,b)$ such that $g’(c)=0$. Furthermore, $g’(x)=f’(x)-\frac{f(b)-f(a)}{b-a}$, so $g’(c)=0$ implies $f’(x)=\frac{f(b)-f(a)}{b-a}$.<div style="text-align: right;">∎</div>

These theorems exemplify the elegance of mathematics. In a cohesive progression, we observe the relationships among these fundamental statements: $\begin{equation} \begin{array}{c} \text{Bolzano-Weierstrass theorem}\\\\ \Downarrow\\\\ \text{Weierestrass's extreme value theorem}\\\\ \Downarrow\\\\ \text{Fermat's theorem}\\\\ \Downarrow\\\\ \text{Roll's theorem}\\\\ \Downarrow\\\\ \text{Mean value theorem} \end{array} \end{equation}$

Strategy of using MVT

Many applications of the mean value theorem (MVT) employ a strategy similar to the one used in proving MVT itself. This strategy typically involves defining a second function, denoted as $h(x)$, which incorporates the function of interest, $f(x)$, along with other variables. Subsequently, MVT is applied to $h(x)$ to derive the desired outcome for $f(x)$. To illustrate this concept, let’s delve into the following example.

Example. Show the following statement: <div id='question-question-calculus1-ch4-proof-mvt7'></div>

proof. <div id='answer-question-calculus1-ch4-proof-mvt7'></div></p><div style="text-align: right;">∎</div>

Of course, the approach outlined above is not the sole way to utilize the Mean Value Theorem. Another valuable application of this theorem is to directly derive useful inequalities. **Example.**

**proof.**

∎

## What does derivatives tell us about the graph of a function ### The first derivative **Definition.** A real function $f$ defined on an interval $I$ is characterized as follows: - If for any $x_1$ and $x_2$ in $I$ where $x_1 < x_2$, it holds that $f(x_1) \leq f(x_2)$, then $f$ is referred to as an increasing function. - If for any $x_1$ and $x_2$ in $I$ where $x_1 < x_2$, it holds that $f(x_1) \geq f(x_2)$, then $f$ is described as a decreasing function. - In the case that $f(x_1) < f(x_2)$ ($f(x_2) > f(x_2)$), where $x_1$ and $x_2$ are in $I$ and $x_1 < x_2$, then $f$ is classified as strictly increasing (decreasing). The determination of whether a function is increasing or decreasing can be made through the use of the first derivative. **Theorem.** A differentiable function $f$ defined on an open interval $I$ satisfies $f'(x) > 0$ for all $x$ in $I$ if and only if $f$ is increasing on $I$. #### Increasing <=> slopes are positive

exp = plot(e^x+100,x,5,10) l1 = plot(e^7*(x-7)+e^7+100,x,6,8,color='red') l2 = plot(e^8*(x-8)+e^8+100,x,7,9,color='red') l3 = plot(e^9*(x-9)+e^9+100,x,8,10,color='red') (exp+l1+l2+l3).show()

#### Decreasing <=> slopes are negative

exp = plot(-(e^x+100),x,5,10) l1 = plot(-e^7*(x-7)-(e^7+100),x,6,8,color='red') l2 = plot(-e^8*(x-8)-(e^8+100),x,7,9,color='red') l3 = plot(-e^9*(x-9)-(e^9+100),x,8,10,color='red') (exp+l1+l2+l3).show()

**The first derivative test.** Suppose $f$ is a differentiable function defined on an open interval $I$, and let $c$ belong to $I$. 1. If $f'(x) < 0$ for $x < c$ in the vicinity of $c$ and $f'(x) > 0$ for $x > c$ in the vicinity of $c$, then $f(c)$ represents a local minimum value. 2. If $f'(x) > 0$ for $x < c$ in the vicinity of $c$ and $f'(x) < 0$ for $x > c$ in the vicinity of $c$, then $f(c)$ represents a local maximum value. Note that the point $c$ in the above test is a critical point of $f$, meaning $f'(c) = 0$. ### The second derivative **Definition.** If $f$ is twice differentiable, the second derivative tells us about the concavity of the graph. Specifically: - If $f''(x) > 0$, the graph is concave upward (convex) at $x$. - If $f''(x) < 0$, the graph is concave down (concave) at $x$. **Second Derivative Test.** Let $f$ be a twice-differentiable function on an open interval $I$ with a critical point at $c$. 1. If $f'(c) = 0$ and $f''(c) > 0$, then $f(c)$ is a local minimum value. 2. If $f'(c) = 0$ and $f''(c) < 0$, then $f(c)$ is a local maximum value. The Second Derivative Test offers a convenient way to determine the concavity of the graph at a critical point and can help identify local extrema with ease. However, it might be inconclusive for some cases, such as functions like $f(x) = x^4$. In such scenarios, we have to revert to using the First Derivative Test to make a definitive determination regarding local extrema. #### Convex <=> slope is increasing

f(x)=x^2+1 @interact def _(n=(5,[3,4,5,6,7,8,9,10])): P=plot(f,(x,-1,1)) fprime=derivative(f,x) points = [-1+2*(k)/n for k in range(n+1)] tangent_lines = [plot(fprime(c)*(x-c)+f(c),x,c-0.5,c+0.5,color="red") for c in points] show(P+sum(tangent_lines))

#### Concave <=> slope is decreasing

f(x)=-(x^2+1) @interact def _(n=(5,[3,4,5,6,7,8,9,10])): P=plot(f,(x,-1,1)) fprime=derivative(f,x) points = [-1+2*(k)/n for k in range(n+1)] tangent_lines = [plot(fprime(c)*(x-c)+f(c),x,c-0.5,c+0.5,color="red") for c in points] show(P+sum(tangent_lines))

If $f(x)$ is differentiable and $f’(x)\neq 1$ for all real numbers $x$, then $f$ has at most one fixed point.

We need to show that if $f$ has fixed points, then $f$ has at most one fixed point. Let $g(x)=f(x)-x$ where $g$ is also continuous and differentiable due to its construction. Then, it is equivalent to show that $g(x)=0$ has at most one solution. If there exists two solutions $a$ and $b$, then there exists a $c$ between $a$ and $b$ such that $g’(c)=0$ by mean value theorem. However, $g’(c)=0$ implies $f’(c)=1$, which contradicts to the assumption. Thus, we can conclude that $f(x)$ has at most one fixed point.

Let $0<a<b$. Show that $\dfrac{b-a}{b}<\ln\dfrac{b}{a}<\dfrac{b-a}{a}.$

Consider the function $f(x)=\ln(x)$ which is continuous and differentiable on $(0,\infty)$. By the mean value theorem, there exists a real number $c$ with $a<c<b$ such that $\ln\left(\frac{b}{a}\right)=\ln(b)-\ln(a)=\frac{1}{c}(b-a).$ Note that $a<c<b$ implies $\frac{1}{a}>\frac{1}{c}>\frac{1}{b}$. Putting everything together, we have the desired conclusion.