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 $\epsilon >0$ such that $f(x)<f(c)$ for all $x\in E\cap \left(c+\epsilon ,c-\epsilon \right)$. Note local because we restrict the function to the $\epsilon$-neighborhood of $c$.

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

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 $f'(c)=0$.

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 $\epsilon >0$ such that for all $\left|h\right|<\epsilon$, $f(c+h)-f(c)\geq 0$.

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

By antisymmetry (in particular, $f'(c)\geq 0$ and $f'(c)\leq 0$) we conclude that $f'(c)=0$.

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

quod erat demonstrandum

Corollary. If a polynomial $P(x)$ has $k>1$ distinct real roots, then the polynomial $P'(x)$ has at least $k-1$ distinct real roots.

Proof. Let $x_{1},\ldots ,x_{k}$ be distinct real roots of $P(x)$ ordered so that $x_{1}<\dots <x_{k}$. By Rolle's Theorem, the derivative $P'(x)$ has a root in each of $k-1$ intervals $(x_{1},x_{2})$, $(x_{2},x_{3})$, ..., $(x_{k-1},x_{k})$.

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Intermediate Value Theorem for Derivatives

Theorem. Suppose that a function $f$ is differentiable on an interval $[a,b]$ with $f'(a)\neq f'(b)$. If $y_{0}$ is a real number between $f'(a)$ and $f'(b)$, then $f'(x_{0})=y_{0}$ for some $x_{0}\in (a,b)$.

Proof. First consider the case when $f'(a)<0$, $f'(b)>0$, 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 $f'(a)<0$, we have $f(a+h)-f(a)<0$ for sufficiently small $h>0$. Hence $x_{0}\neq a$. Similarly, $f'(b)>0$ implies that $f(b+h)-f(b)<0$ for sufficiently small $h<0$. Hence $x_{0}\neq b$. We obtain that $x_{0}\in (a,b)$. Then $f'(x_{0})=0$ due to Fermat's theorem.

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

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

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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 $f'$ 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 $\left[a,b\right]$ and differentiable on $\left(a,b\right)$, then there is a $c\in (a,b)$ such that $f(b)-f(a)=f'(c)\,(b-a)$.

In other words, there exists a point $c$ between two endpoints $a$ and $b$ such that the derivative $f'(c)$ (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, ${h_{0}}'(x)={\frac {f(b)-f(a)}{b-a}}$ 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, $h'(c)=0$ for some $c\in (a,b)$. We have $h'(c)=f'(c)-h_{0}'(c)=f'(c)-{\frac {f(b)-f(a)}{b-a}}$, thus $f'(c)={\frac {f(b)-f(a)}{b-a}}$ or equivalently, $f(b)-f(a)=f'(c)\,(b-a)$.

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 $f'\geq 0$ on $(a,b)$.
2. $f$ is decreasing on $[a,b]$ if and only if $f'\leq 0$ on $(a,b)$.
3. $f$ is strictly increasing on $[a,b]$ if (not biconditional) $f'>0$.
4. $f$ is strictly decreasing on $[a,b]$ if (not biconditional) $f'<0$.
5. $f$ is constant on $[a,b]$ if and only if $f'=0$ on $(a,b)$.

Proof. Let $a\leq x_{1}<x_{2}\leq b$. By the mean value theorem, $f(x_{2})-f(x_{1})=f'(c)\,(x_{2}-x_{1})$ for some $c\in (x_{1},x_{2})$. Obviously $f'(c)>0$ if and only if $f(x_{1})<f(x_{2})$. Likewise, $f'(c)\geq 0$ if and only if $f(x_{1})\leq f(x_{2})$. 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 $\lim _{x_{1}\to x_{2}}{\frac {f(x_{1})-f(x_{2})}{x_{1}-x_{2}}}$) 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

$g'(c)\,\left(f(b)-f(a)\right)=f'(c)\,\left(g(b)-g(a)\right)$

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 $h(x)=f(x)\,\left(g(b)-g(a)\right)-g(x)\,\left(f(b)-f(a)\right)$. 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, $h'(c)=0$ for some $c\in (a,b)$. It remains to notice that $h'(c)=f'(c)\,\left(g(b)-g(a)\right)-g'(c)\,\left(f(b)-f(a)\right)$.

Therefore $g'(c)\,\left(f(b)-f(a)\right)=f'(c)\,\left(g(b)-g(a)\right)$

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Taylor's Formula

Theorem. Let $n\in \mathbb {N}$ 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 $x_{0}\in I$, $x_{0}\neq x$, there exists a point between $x_{0}$ and $x$ such that $G'(c)\,\left(F(x)-F(x_{0})\right)=F'(c)\,\left(G(x)-G(x_{0})\right)$. 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

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

Summing up over $k$ from $1$ to $n$, we obtain that $G'(t)=-{\frac {f^{(n+1)}(t)}{n!}}\,(x-t)^{n}$. Finally, $F'(t)=-{\frac {(x-t)^{n}}{n!}}$ so that ${\frac {G'(t)}{F'(t)}}=f^{(n+1)}(t)$ for $t\neq x$. 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 $P_{n}^{f,x_{0}}(x){=}f(x_{0})+{\frac {f'(x_{0})}{1!}}\,(x-x_{0})+\dots +{\frac {f^{(n)}(x_{0})}{n!}}\,(x-x_{0})^{n}$ 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 $1\leq k\leq n$.