04-ApplicationsOfDifferentiation.Rmd

#  (PART)  Applications of Differentiation {-}



#  4.1  Maximum and Minimum Values {-}

##  Critical Points {-}

In this section, we examine the idea of critical points.

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**Definition:**  A point $x = c$ is called a critica point of a function $f$ if $f(c)$ exists and either \[f^{\prime}(c) = 0 \hspace{1in} \text{or} \hspace{1in} f^{\prime}(c) \text{does not exist}.\]
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The fact that $f(c)$ exists  means that $x=c$ is in the domain of the function and is therefore an actual point.


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**Example:**    Determine the critical points of $f(x) = \dfrac{3}{20}x^{4/3}(2x^2-5)$.
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**Solution:**    We begin by differentiating $f$ using the product rule.  Let $u = \dfrac{3}{20}x^{4/3}$ and $v = 2x^2-5$.  Then $u^{\prime} = \dfrac{1}{5}x^{1/3}$ and $v^{\prime} = 4x$.  So,
\begin{align*} f^{\prime}(x) &= u^{\prime}v + v^{\prime}u\\
&= \dfrac{1}{5}x^{1/3}(2x^2-5) + 4x\left(\dfrac{3}{20}x^{4/3} \right)\\
&= \dfrac{1}{5}x^{1/3}\left(2x^2-5 + 3x^2\right)\\
&= \dfrac{1}{5}x^{1/3}\left(5x^2-5\right)\\
&= x^{1/3}\left(x^2-1\right). \end{align*}
So, \[f^{\prime}(x) = \sqrt[3]{x}(x^2-1).\]  We find that $f^{\prime}(x) = 0$ when $x = 0, \pm 1$.  All three of those values are in the domain of $f$, so all three of those values are critical points.  We also note that $f^{\prime}(x)$ exists everywhere, so there are no critical points where $f^{\prime}(x)$ does not exist.
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**Example:**    Determine the critical points of $f(x) = \dfrac{1}{5} \left(2\ln(x-3) + 3\ln(x+2) \right)$.
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**Solution:**    We begin by differentiating $f$ .
\begin{align*} f^{\prime}(x) &= \dfrac{1}{5} \left(\dfrac{2}{x-3} + \dfrac{3}{x+2} \right)\\
&= \dfrac{1}{5} \left(\dfrac{2(x+2)}{x-3} + \dfrac{3(x-3)}{x+2} \right)\\
&= \dfrac{2x+4+3x-9}{5(x+2)(x-3)} \\
&= \dfrac{5x-5}{5(x+2)(x-3)} \\
&= \dfrac{x-1}{(x+2)(x-3)}. \end{align*}
So, \[f^{\prime}(x) = \dfrac{x-1}{(x+2)(x-3)}.\]  We find that $f^{\prime}(x) = 0$ when $x = 1$ and that $f^{\prime}(x)$ does not exist when $x = -2$ and $x=3$.  These are the three possible critical points of $f$.  However, we note that none of these points are in the domain of $f$ and therefore none are critical points.  Therefore, $f$ has no critical points.
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**Example:**    Determine the critical points of $f(x) = 3x - 2\cos(3x)$.
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**Solution:**   We begin by differentiating $f$.  \[f^{\prime}(x) = 3 +6\sin(3x) = 3(1+2\sin(3x)).\]  So, $f^{\prime}(x) = 0$ when $1 + 2\sin(3x) = 0$ or when $\sin(3x) = -\dfrac{1}{2}$.  Of course, this happens when $3x = -\pi/6 + 2k\pi$ for $k \in \mathbb{Z}$.  Simplifying, we see that $x = -\pi/18 + 2/3 k \pi$ for $k \in \mathbb{Z}$.  There are no points where $f^{\prime}(x)$ is undefined.  Since $f$ is defined for all $x \in \mathbb{R}$, the set of critical points for $f$ is $x = -\pi/18 + 2/3 k \pi$, with $k \in \mathbb{Z}$.
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Practice Problems

1.  Determine the critical points for each of the following functions.

  *  $f\left( x \right) = 2{x^5} + 11{x^4} - 10{x^3} + 23$
  *  $f\left( x \right) = \dfrac{x^2+1}{x^2-x-6}$
  *  $f\left( x \right) = 2xe^{8-x^2}$
  *  $f\left( x \right) = x^3\ln(2x)$
  *  $f\left( x \right) = 4\cos(x) - x$
  *  $f\left( x \right) = \sin^2(x) - \sin^4(x)$


2.  Identify the critical points on the graph below.

|![](Pic48.png)|
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3.  Identify the critical points on the graph below.

|![](Pic49.png)|
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##  Maximum and Minimum Values {-}

Although it is fairly simple to identify maximum and minimum values on a graph, a more rigorous definition is required.

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**Definition:** 

  *  We say that $f(x)$ has an global (or absolute) maximum at $x=c$ if $f(x) \leq f(c)$ for every $x$ in the domain on which we are working.\index{absolute maximum}
  *  We say that $f(x)$ has a local (or relative) maximum at $x=c$ if $f(x) \leq f(c)$ for every $x$ in some open interval around $x= c$.\index{relative maximum}\index{local maximum}
  *  We say that $f(x)$ has an global (or absolute) minimum at $x=c$ if $f(x) \geq f(c)$ for every $x$ in the domain on which we are working.\index{absolute minimum}\index{global minimum}
  *  We say that $f(x)$ has a relative (or local) minimum at $x=c$ if $f(x) \geq f(c)$ for every $x$ in some open interval around $x= c$.
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|![](Pic50.png)|
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Alternatively, we call maximum and minimum points of a function where local extrema refer to local maximum and minimum values and global extrema refer to global maximum and minimum values.



Another important note is that for a point $x=c$ to be a relative extrema, there must be $x$-values to the left and right of $c$.  For example, a relative minimum cannot occur at the endpoint of an interval.


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**Example:**  Identify the local and global extrema of $ y = x^2$ on $[-2,3]$.
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**Solution:**   We plot $y = x^2$ and see that $(0,0)$ is the local and global minimum values and $(3,9)$ is the global maximum value.  Note that there is no local maximum value.
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|![](Pic51.png)|
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If we change the interval for $y = x^2$,  then the maximum and minimum values may also change.  For example, if the interval for $y = x^2$ is $[-3,3]$, there are two global maximum values - one at -3 and one at 3.  If the interval for $y=x^2$ is $\mathbb{R}$ or another open interval, there is no maximum value.  Of course, if we examined a function like $y = x$ on $\mathbb{R}$ or on an open interval, we get no maximum or minimum values.  These observations lead us to the following theorem.

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**Theorem (Extreme Value Theorem):**   If $f$ is continuous on the closed interval $[a,b]$, there are two numbers $c$ and $d$ with $a \leq c, d \leq b$ so that $f(c)$ is an absolute maximum for $f$ and $f(d)$ is an absolute minimum for $f$.
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**Example:**   Find the extreme values  of $y = -(x-3)^3$ on $[-1,5]$.
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**Solution:**   We sketch the graph of $y$ and notice that the extreme values are a maximum at $x=-1$ and a minimum value at $x = 5$.
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|![](Pic52.png)|
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It is worth noting that if a function is not continuous at a point on $[a,b]$, it does not mean that the function does not have extrema.



There is a link between local extrema and critical points.  

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**Theorem (Fermat's Theorem):**   If $f$ has a local extrema at $x = c$ and $f^{\prime}(c)$ exists, then $x = c$ is a critical point of $f(x)$.
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**Example:**    Use Fermat's theorem to find the extrema of $y = x^2 - 4x$.
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**Solution:**   We differentiate $y$ to find \[y^{\prime} = 2x - 4.\]  Critical points occur where $y^{\prime} = 0$ or where $x = 2$ in this case.  Therefore, $x = 2$ is the only extreme value.  Since the $y$-value here is -4 and $y$-values close to $x=2$ are larger (ie, at $x=1$, we have $y = -2$), we see that $x = 2$ is a local minimum.
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Practice Problems

1.  Determine the extreme values for the following graphs on the given interval.

  * $f(x) = 5 - 2x$ on $[-2,2]$
  * $f(x) = (x-3)^2$ on $[-1,4]$
  * $f(x) = e^{x-3}$ on $[-3,5]$
  * $f(x) = \sin(x)$ on $[-\pi,\pi]$


2.  Identify the critical points on the graph below.


|![](Pic53.png)|
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\noindent 3.  Identify the critical points on the graph below.


|![](Pic54.png)|
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##  Proof of the Extreme Value Theorem {-}

In this section, we prove the Extreme Value Theorem and Fermat's Theorem.  Prior to these proofs, we prove a preliminary result.

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**Theorem (Boundedness Theorem):**   If $f$ is continuous on $[a,b]$, then it is bounded on $[a,b]$.
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**Proof:**  Consider the set $B$ of points $x in  [a,b]$ such that $f(x)$ is bounded on $[a,x]$. We note that $a$ is one such point, for $f(x)$ is bounded on $[a,a]$ by the value $f(a)$. If $e>a$ is another point, then all points between $a$ and $e$ also belong to $B$. In other words $B$ is an interval closed at its left end by $a$.

Now $f$ is continuous on the right at $a$, hence there exists $\delta >0$ such that $|f(x)-f(a)|<1$ for all $x \in [a,a+\delta ]$. Thus $f$ is bounded by $f(a)-1$ and $f(a)+1$ on the interval$[a,a+\delta ]$ so that all these points belong to $B$.

So far, we know that $B$ is an interval of non-zero length, closed at its left end by $a$.

Next,$B$ is bounded above by $b$. Hence the set $B$ has a supremum in $[a,b]$; let us call it $s$. From the non-zero length of $B$ we can deduce that $s>a$.

Suppose $s<b$. Now $f$ is continuous at $s$, hence there exists $\delta >0$ such that $|f(x)-f(s)|<1$ for all $x \in [s-\delta ,s+\delta ]$ so that $f$ is bounded on this interval. But it follows from the supremacy of $s$ that there exists a point belonging to $B$, $e$ say, which is greater than $s-\delta /2$. Thus $f$ is bounded on $[a,e]$ which overlaps $[s-\delta ,s+\delta ]$ so that $f$ is bounded on $[a,s+\delta ]$. This however contradicts the supremacy of $s$.

We must therefore have $s=b$. Now $f$ is continuous on the left at $s$, hence there exists $\delta >0$ such that $|f(x)-f(s)|<1$ for all $x \in [s-\delta ,s]$ so that $f$ is bounded on this interval. But it follows from the supremacy of $s$ that that there exists a point belonging to $B$, $ e$ say, which is greater than $s-\delta /2$. Thus $f$ is bounded on $[a,e]$ which overlaps $[s-\delta ,s]$ so that $f$ is bounded on $[a,s]$.
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**Theorem (Extreme Value Theorem):**   If $f$ is continuous on the closed interval $[a,b]$, there are two numbers $c$ and $d$ with $a \leq c, d \leq b$ so that $f(c)$ is an absolute maximum for $f$ and $f(d)$ is an absolute minimum for $f$.
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**Proof:** The set $\{y \in R : y = f(x) \text{ for some } x \in [a,b]\}$ is a bounded set. Hence, its least upper bound exists by least upper bound property of the real numbers. Let $M = \sup(f(x))$ on $[a, b]$. If there is no point $x$ on $[a, b]$ so that $f(x) = M$ then $f(x) < M$ on $[a, b]$. Therefore, $1/(M - f(x))$ is continuous on $[a, b]$.

However, to every positive number $\varepsilon$, there is always some $x \in [a, b]$ such that $M - f(x) < \varepsilon$ because $M$ is the least upper bound. Hence, $1/(M - f(x)) > 1/\varepsilon$, which means that $1/(M - f(x))$ is not bounded. Since every continuous function on a $[a, b]$ is bounded, this contradicts the conclusion that $1/(M - f(x))$ was continuous on $[a, b]$. Therefore, there must be a point $x \in [a, b]$ such that $f(x) = M$.
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**Theorem (Fermat's Theorem):**   If $f$ has a local extrema at $x = c$ and $f^{\prime}(c)$ exists, then $x = c$ is a critical point of $f(x)$.
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**Proof:** Without loss of generality, assume that $f(x)$ has a relative maximum at $x = c$. We know that $f(c) \geq f(x)$ for all $x$ sufficiently close to $x = c$.  In particular for all $h$ that are sufficiently close to zero (positive or negative) we must have, \[f(c) \geq f(c+h),\] or \[f(c+h) - f(c) \leq 0.\]

Now, at this point, assume  $h>0$.  Then \[\dfrac{f(c+h)-f(c)}{h} \leq 0.\]  Since $h > 0$,  we can take the right-hand limit of both sides and find \[\lim_{h \rightarrow 0^+} \dfrac{f(c+h)-f(c)}{h} \leq \lim_{h\rightarrow 0^+} 0 = 0.\]  Since $f^{\prime}(c)$ exists, it must be true that $f^{\prime}(c) \leq 0$.

If $h < 0$, \[\dfrac{f(c+h)-f(c)}{h} \geq 0.\]  For the same reason as earlier, we find that \[f^{\prime}(c) = \lim_{h \rightarrow 0} \dfrac{f(c+h)-f(c)}{h}= \lim_{h \rightarrow 0^-} \dfrac{f(c+h)-f(c)}{h} \geq \lim_{h \rightarrow 0^-} 0 = 0.\]  Since $f^{\prime}(c) \geq 0$ and $f^{\prime}(c) \leq 0$, it must be true that $f^{\prime}(c) = 0$ and so $c$ is a critical point.

The proof for a relative minimum is similar.
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Practice Problems

1.  Sketch the graph of some function on the interval $[1,6]$ that has an absolute maximum at $x=6$ and an absolute minimum at $x=3$.

2.  Sketch the graph of some function on the interval $[-4,3]$ that has an absolute maximum at $x=-3$ and an absolute minimum at $x=2$.

3.  Sketch the graph of some function that meets the following conditions:

  *  The function is continuous.
  *  Has two relative minimums.
  *  One of relative minimums is also an absolute minimum and the other relative minimum is not an absolute minimum.
  *  Has one relative maximum.
  *  Has no absolute maximum.


4.  Sketch the graph of some function that meets the following conditions:

  *  Graphed on the interval $[4,10]$.
  *  Has no relative extrema.
  *  Has an absolute maximum at one end point.
  *  Has an absolute minimum at the other end point.

5.  Sketch the graph of some function that meets the following conditions:

  *  Has a discontinuity at some point.
  *  Has an absolute maximum and an absolute minimum.
  *  Neither absolute extrema occurs at the discontinuity.

6.  Sketch the graph of a function on the interval $(-\infty,\infty)$ that has a relative minimum at $x=4$ and a relative maximum at $x = -2$ and no absolute extrema.

7.  Sketch the graph of some function that meets the following conditions:

  *  Has at least one absolute maximum.
  *  Has one relative minimum.
  *  Has no absolute minimum.




##  Finding Absolute Extrema {-}

Given a continuous function $f$ on a closed interval $[a,b]$, the extreme value theorem guarantees both a global maximum and global minimum value for $f$.  The process for finding these extrema is quite simple:

  *  Verify the conditions of the extreme value theorem are met - ie, is $f$ continuous on a closed interval $[a,b]$.
  *  Find all critical points of $f$ on $[a,b]$ - Fermat's Theorem  guarantees that these will be the local extrema.
  *  Evaluate $f$ at each critical point and at both $a$ and $b$.
  *  Determine the global extrema.




\begin{example}  Find the global maximum and minimum for $p(x) = 3x^2 - 12x + 5$ on $[0,3]$.\\

\noindent\textbf{Solution:} First, we note that $p$ is a polynomial and is therefore continuous on $\mathbb{R}$.  The derivative of $p$ is \[p^{\prime}(x) = 6x - 12 = 6(x-2).\]  Of course, $p^{\prime}(x) = 0$ at $x = 2$.  That is the only critical point of $p$.  We now evaluate $p$ at the critical point ($x=2$) and the two end points of the interval ($x = 0,3$).  Since \[p(0) = 5, \hspace{0.5in} p(2) = -7, \hspace{0.5in} p(3) = -4,\] the global maximum is at $x=0$ and the global minimum is at $x=2$.
\end{example}




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**Example:**  Find the global maximum and minimum for $f(x) = \dfrac{x}{x^2+1}$ on $[0,5]$.
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**Example:**  First, we note that $f$ is a rational function and is therefore continuous on its domain (in this case $\mathbb{R}$).  The derivative of $f$ is \[f^{\prime}(x) =\dfrac{x^2+1 - 2x^2}{(x^2+1)^2} = \dfrac{1- x^2}{(x^2+1)^2}.\]  Of course, $f^{\prime}(x) = 0$ at $x = \pm 1$.  The only critical point of $f$ on $[0,5]$ is $x=1$.  We now evaluate $f$ at the critical point ($x=1$) and the two end points of the interval ($x = 0,5$).  Since \[f(0) = 0, \hspace{0.5in} f(1) = 1/2, \hspace{0.5in} f(5) = 5/26,\] the global maximum is at $x=1$ and the global minimum is at $x=0$.
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**Example:**  Find the global maximum and minimum for $f(x) = x\sqrt{9-x^2}$ on $[-1,3]$.
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**Example:**  First, we note that $f$ is continuous on $(-\infty,3]$.  The derivative of $f$ is \[f^{\prime}(x) =\sqrt{9-x^2} + \dfrac{-2x^2}{2\sqrt{9-x^2}} = \dfrac{-2x^2+9}{\sqrt{9-x^2}}.\]  Of course, $f^{\prime}(x) = 0$ at $x = \pm 3\sqrt{2}/2$.  The only critical point of $f$ on $[-1,3]$ is $x=3\sqrt{2}/2$.  We now evaluate $f$ at the critical point ($x=3\sqrt{2}/2$) and the two end points of the interval ($x = -1,3$).  Since \[f(-1) = -2\sqrt{2}, \hspace{0.5in} f(3\sqrt{2}/2) = 9/2, \hspace{0.5in} f(3) = 0,\] the global maximum is at $x=3\sqrt{2}/2$ and the global minimum is at $x=-1$.
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Practice Problems

1.  Determine the extreme values for the following graphs on the given interval.

  *  $f(x) = x^4 - 2x^2 + 4$ on $[-2,3]$
  *   $f(x) = x-2\cos(x)$ on $[-\pi,\pi]$
  *  $f(x) = x\sqrt{4-x^2}$ on $[-1,2]$
  *  $f(x) = \dfrac{3-4x}{x^2+1}$ on $[-2,4]$
  *  $f(x) = 8 + 3x + 7{x^2} - {x^3}$ on $[-1,5]$
  *  $f(x) = 3\cos \left( {2x} \right) - 5x$ on $[0,6]$
  *   $f(x) = 7\cos \left( x \right) + 2x$ on $[-5,4]$
  *  $f(x) = e^{5 + 9x} + e^{1 - 3x} + 6$ on $[-1,0]$