MATH 409 Lecture 16

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Lecture Slides

Mean Value Theorem

Points of Local Extremum

Let $f:E\to \mathbb{ℝ}$ be a function defined on $E\subset \mathbb{ℝ}$.

We say that $f$ attains a local maximum at a point $c\in E$ if there exists $ϵ>0$ such that $f(x)<f(c)$ for all $x\in E\cap (c+ϵ,c-ϵ)$. Note local because we restrict the function to the $ϵ$-neighborhood of $c$.

Similarly, $f$ attains a local minimum at a point $c\in E$ if there exists $ϵ>0$ such that $f(x)\ge f(c)$ for all $x\in E\cap (c-ϵ,c+ϵ)$.

Fermat's Theorem

not little or last...

Theorem. [Fermat.] If a function $f$ is differentiable at a point $c$ of local extremum (maximum or minimum), then $f^{\prime }(c)=0$.

Proof. Assume without loss of generality that $c$ is a point of local minimum. Since $f$ is defined on an open interval containing $c$, there exists $ϵ>0$ such that for all $\left|h\right|<ϵ$, $f(c+h)-f(c)\ge 0$.

1. In the case $h>0$, this implies $\underset{h\to 0^{+}}{\lim }\frac{f(c+h)-f(c)}{h}\ge 0$
2. In the case $h<0$, this implies $\underset{h\to 0^{-}}{\lim }\frac{f(c+h)-f(c)}{h}\le 0$

By antisymmetry (in particular, $f^{\prime }(c)\ge 0$ and $f^{\prime }(c)\le 0$) we conclude that $f^{\prime }(c)=0$.

quod erat demonstrandum

Rolle's Theorem

Theorem. Suppose that $a,b\in \mathbb{ℝ}$ with $a<b$. If a function $f$ is continuous on the interval $[a,b]$, differentiable on $(a,b)$, and if $f(a)=f(b)$, then $f^{\prime }(c)=0$ for some $c\in (a,b)$.

Proof. By the Extreme Value Theorem, the function $f$ attains its (absolute) maximum $M$ and minimum $m$ on $[a,b]$. We consider two cases:

1. In the case $M\ne m$, at least one of the extrema is attained at a point $c$ in $(a,b)$. Then $f^{\prime }(c)=0$ by Fermat's theorem.
2. In the case $M=m$, the function $f$ is constant on $[a,b]$. Then $f^{\prime }(c)=0$ for all $c\in (a,b)$.

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Corollary. If a polynomial $P(x)$ has $k>1$ distinct real roots, then the polynomial $P^{\prime }(x)$ has at least $k-1$ distinct real roots.

Proof. Let $x_1,\dots ,x_k$ be distinct real roots of $P(x)$ ordered so that $x_1<\dots <x_k$. By Rolle's Theorem, the derivative $P^{\prime }(x)$ has a root in each of $k-1$ intervals $(x_1,x_2)$, $(x_2,x_3)$, ..., $(x_{k-1},x_k)$.

quod erat demonstrandum

Intermediate Value Theorem for Derivatives

Theorem. Suppose that a function $f$ is differentiable on an interval $[a,b]$ with $f^{\prime }(a)\ne f^{\prime }(b)$. If $y_0$ is a real number between $f^{\prime }(a)$ and $f^{\prime }(b)$, then $f^{\prime }(x_0)=y_0$ for some $x_0\in (a,b)$.

Proof. First consider the case when $f^{\prime }(a)<0$, $f^{\prime }(b)>0$, and $y_0=0$. Since $f$ is differentiable on $[a,b]$, it must be continuous on $[a,b]$. By the extreme value theorem, $f$ attains its absolute minimum on $[a,b]$ at some point $x_0$. Since $f^{\prime }(a)<0$, we have $f(a+h)-f(a)<0$ for sufficiently small $h>0$. Hence $x_0\ne a$. Similarly, $f^{\prime }(b)>0$ implies that $f(b+h)-f(b)<0$ for sufficiently small $h<0$. Hence $x_0\ne b$. We obtain that $x_0\in (a,b)$. Then $f^{\prime }(x_0)=0$ due to Fermat's theorem.

Now consider $f^{\prime }(a)>0$, $f^{\prime }(b)<0$, and $y_0=0$. Then the function $g=-f$ is differentiable on $[a,b]$ with $g^{\prime }(a)=-f^{\prime }(a)<0$ and $g^{\prime }(b)=-f^{\prime }(b)>0$. By the previous case, $g^{\prime }(x_0)=0$ for some $x_0\in (a,b)$. Then $f^{\prime }(x_0)=-0=0$.

In the general case when $y_0\ne 0$, consider a function $h(x)=f(x)-y_{0}x$. It is differentiable on $[a,b]$ and $h^{\prime }(x)=f^{\prime }(x)-y_0$ for all $x\in [a,b]$. It follows that $0$ lies between $h^{\prime }(a)$ and $h^{\prime }(b)$. By the above, $h^{\prime }(x_0)=0$ for some $x_0\in (a,b)$. Then $f^{\prime }(x_0)=h^{\prime }(x_0)+y_0=y_0$.

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Note: We had a similar theorem about continuous functions. Now although $f$ is continuous on $[a,b]$, its derviative need not be continuous.

Corollary. If a function $f$ is differentiable on an open interval $(c,d)$, then the derivative $f^{\prime }$ has no jump discontinuities in $(c,d)$.

If a derivative has a jump discontinuity, then there are many values $y$ for which there can be no $x$ such that $f(x)=y$. This means that not every function can be a derivative.

Mean Value Theorem

Theorem. If a function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there is a $c\in (a,b)$ such that $f(b)-f(a)=f^{\prime }(c)(b-a)$.

In other words, there exists a point $c$ between two endpoints $a$ and $b$ such that the derivative $f^{\prime }(c)$ (i.e. slope of the tangent line) is equivalent to the slope of the chord line through $a$ and $b$ $\frac{f(b)-f(a)}{b-a}$.

Proof. Let $h_0(x)=\frac{f(a)(b-x)+f(b)(x-a)}{b-a}$ for $x\in \mathbb{ℝ}$.

We observe that the function $h_0$ is differentiable. Moreover, $h_0\text{'}(x)=\frac{f(b)-f(a)}{b-a}$ for all $x\in \mathbb{ℝ}$. By construction, $h_0(a)=f(a)$ and $h_0(b)=f(b)$.

It follows that the function $h=f-h_0$ is continuous on $[a,b]$, differentiable on $(a,b)$, and satisfies $h(a)=h(b)=0$. By Rolle's Theorem, $h^{\prime }(c)=0$ for some $c\in (a,b)$. We have $h^{\prime }(c)=f^{\prime }(c)-{h_0}^{\prime }(c)=f^{\prime }(c)-\frac{f(b)-f(a)}{b-a}$, thus $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$ or equivalently, $f(b)-f(a)=f^{\prime }(c)(b-a)$.

quod erat demonstrandum

Monotone Functions (Revisited)

Theorem. Suppose that a function $f$ is continuous on an interval $[a,b]$ and differentiable on $(a,b)$.

1. $f$ is increasing on $[a,b]$ if and only if $f^{\prime }\ge 0$ on $(a,b)$.
2. $f$ is decreasing on $[a,b]$ if and only if $f^{\prime }\le 0$ on $(a,b)$.
3. $f$ is strictly increasing on $[a,b]$ if (not biconditional) $f^{\prime }>0$.
4. $f$ is strictly decreasing on $[a,b]$ if (not biconditional) $f^{\prime }<0$.
5. $f$ is constant on $[a,b]$ if and only if $f^{\prime }=0$ on $(a,b)$.

Proof. Let $a\le x_1<x_2\le b$. By the mean value theorem, $f(x_2)-f(x_1)=f^{\prime }(c)(x_2-x_1)$ for some $c\in (x_1,x_2)$. Obviously $f^{\prime }(c)>0$ if and only if $f(x_1)<f(x_2)$. Likewise, $f^{\prime }(c)\ge 0$ if and only if $f(x_1)\le f(x_2)$. This proves statements 3 and 4, and the (⇒) part of statements 1 and 2. The (⇐) parts of 1 and 2 follows from the Comparison Theorem (by taking $\underset{x_1\to x_2}{\lim }\frac{f(x_1)-f(x_2)}{x_1-x_2}$) Finally, statement 5 follows from antisymmetry between 1 and 2.

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Generalized Mean Value Theorem

Theorem. If $f$ and $g$ are continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a $c\in (a,b)$ such that

$g^{\prime }(c)(f(b)-f(a))=f^{\prime }(c)(g(b)-g(a))$

If either $g(x)$ or $f(x)$ is the identity function, we get exactly the mean value theorem.

Proof. For any $x\in [a,b]$, let $h(x)=f(x)(g(b)-g(a))-g(x)(f(b)-f(a))$. This is simply a linear combination of $f$ and $g$. Observe that the function $h$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Further, $h(a)=f(a)g(b)-g(a)f(b)$ and $h(b)=f(a)g(b)-g(a)f(b)$. Hence $h(a)=h(b)$. By Rolle's Theorem, $h^{\prime }(c)=0$ for some $c\in (a,b)$. It remains to notice that $h^{\prime }(c)=f^{\prime }(c)(g(b)-g(a))-g^{\prime }(c)(f(b)-f(a))$.

Therefore $g^{\prime }(c)(f(b)-f(a))=f^{\prime }(c)(g(b)-g(a))$

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Taylor's Formula

Theorem. Let $n\in \mathbb{\mathbb{N}}$ and $I\subset \mathbb{ℝ}$ be an open interval. If a function $f:I\to \mathbb{ℝ}$ is $n+1$ times differentiable on $I$, then for each pair of points $x,x_0\in I$, there is a point $c$ between $x$ and $x_0$ such that

$f(x)=f(x_0)+\sum _{k=1}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k+\frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0{)}^{n+1}$

Proof. Let us fix $x\in I$ and define

$F(t)=\frac{(x-t{)}^{n+1}}{(n+1)!}$ and $G(t)=f(x)-f(t)-\sum _{k=1}^n\frac{f^{(k)}(t)}{k!}(x-t{)}^k$

The function $F$ is infinitely differentiable on $\mathbb{ℝ}$. The function $G$ is defined and differentiable on $I$. By the Generalized Mean Value Theorem, for every $x_0\in I$, $x_0\ne x$, there exists a point between $x_0$ and $x$ such that $G^{\prime }(c)(F(x)-F(x_0))=F^{\prime }(c)(G(x)-G(x_0))$. Note that the latter follows both in the case $x_0<x$ and in the case $x_0>x$ (both sides of the equation get negated, so it remains equivalent). Clearly $F(x)=G(x)=0$. Further

$\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{f^{(k)}(t)}{k!}{(x-t)}^k\right)=\frac{f^{(k+1)}(t)}{k!}(x-t{)}^k-\frac{f^{(k)}(t)}{(k-1)!}(x-t{)}^{k-1}$

Summing up over $k$ from $1$ to $n$, we obtain that $G^{\prime }(t)=-\frac{f^{(n+1)}(t)}{n!}(x-t{)}^n$. Finally, $F^{\prime }(t)=-\frac{(x-t{)}^n}{n!}$ so that $\frac{G^{\prime }(t)}{F^{\prime }(t)}=f^{(n+1)}(t)$ for $t\ne x$. It follows that $G(x_0)=f^{(n+1)}(c)F(x_0)$, which implies Taylor's formula.

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Note: The function $P_n^{f,x_0}(x)=f(x_0)+\frac{f^{\prime }(x_0)}{1!}(x-x_0)+\dots +\frac{f^{(n)}(x_0)}{n!}(x-x_0{)}^n$ is a polynomial of degree at most $n$. It is called the Taylor polynomial of order $n$ generated by $f$ centered at $x_0$. One can check that $P_n^{f,x_0}(x_0)=f(x_0)$ and $(P_n^{f,x_0}{)}^{(k)}(x_0)=f^{(k)}(x_0)$ for $1\le k\le n$.