Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 5

Applications of the Derivative

Local and Absolute Extreme Values

The intervals of decrease and increase of a function can be used to determine the local extreme points; that is, the points where the function has a local maximum or minimum. The graph in Figure 5.11, below, shows the local extreme points of a function $f$.

Figure 5.11. Local extreme points of a function $f$

Example 5.28. From the basic graphs of functions in Table 2, $f(x) = {x^2}$ and $g(x) = |x|$ have only one local minimum, at ${\displaystyle 0}$, and that ${\displaystyle f\left(0\right)=0=g\left(0\right)}$ is the local minimum value.

The sine function has infinitely many local maxima and minima, at

${\displaystyle {\displaystyle \frac{\left(4k+1\right)\pi }{2}}\text{and}{\displaystyle \frac{\left(4k-1\right)\pi }{2}}\text{,}}$

respectively, for any integer $k$. The local minimum value is ${\displaystyle -}$1 and the local maximum value is ${\displaystyle 1}$.

The functions ${\displaystyle h\left(x\right)=x^3\text{,}}$ tangent and cotangent do not have local extreme points.

Example 5.29. In Example 5.26, we found the intervals of decrease and increase of the function

${\displaystyle y=x^3-6x^2-15x+4\text{.}}$

The function has a local maximum at ${\displaystyle -1}$ with local maximum value of ${\displaystyle y\left(-1\right)=12}$, and a local minimum at ${\displaystyle 5}$ with minimum local value of ${\displaystyle y\left(5\right)=-96}$.

Example 5.30. The function

${\displaystyle y={\displaystyle \frac{x}{x^2+4x+3}}}$

of Example 5.27 has a relative minimum at ${\displaystyle -\sqrt{3}}$ with relative minimum value of

${\displaystyle f\left(-\sqrt{3}\right)={\displaystyle \frac{-\sqrt{3}}{6-4\sqrt{3}}}\text{.}}$

Although the function changes from increasing to decreasing at ${\displaystyle -3}$, the function does not have a relative maximum at this point, because it is not defined at this point.

From Figure 5.11, we can see that the following result is plausible, and we will prove it.

Theorem 5.8. If a function $f$ has a local maximum or local minimum at $c$, then $c$ is a critical number of $f$, that is either ${\displaystyle f^{\prime }\left(c\right)=0}$ or ${\displaystyle f^{\prime }\left(c\right)}$ does not exist.

Proof. If we assume that ${\displaystyle f^{\prime }\left(c\right)}$ exists, then we must explain why ${\displaystyle f^{\prime }\left(c\right)=0}$.

By Definition 5.7, above, if $f$ has a minimum at $c$, then ${\displaystyle f\left(c\right)\le f\left(c+h\right)}$ for any $h$ such that ${\displaystyle \left|h\right|<\delta }$ for some ${\displaystyle \delta >0}$. So, for this $h$,

${\displaystyle f\left(c+h\right)-f\left(c\right)\ge 0}$

Observe that ${\displaystyle c+h}$ is near $c$ from the right if ${\displaystyle h>0}$ and from the left if ${\displaystyle h<0}$. Hence,

${\displaystyle {\displaystyle \frac{f\left(c+h\right)-f\left(c\right)}{h}}\ge 0\text{ for }h>0\text{and}{\displaystyle \frac{f\left(c+h\right)-f\left(c\right)}{h}}\le 0\text{ for }h<0\text{.}}$

Then, by Theorem 3.43, we conclude that

${\displaystyle \underset{h\to 0^{+}}{lim}{\displaystyle \frac{f\left(c+h\right)-f\left(c\right)}{h}}\ge \underset{h\to 0^{+}}{lim}0=0}$

and

${\displaystyle \underset{h\to 0^{-}}{lim}{\displaystyle \frac{f\left(c+h\right)-f\left(c\right)}{h}}\le \underset{h\to 0^{-}}{lim}0=0}$

Since ${\displaystyle f^{\prime }\left(c\right)}$ exists,

${\displaystyle f^{\prime }\left(c\right)=\underset{h\to 0^{+}}{lim}{\displaystyle \frac{f\left(c+h\right)-f\left(c\right)}{h}}=\underset{h\to 0^{-}}{lim}{\displaystyle \frac{f\left(c+h\right)-f\left(c\right)}{h}}=0\text{.}}$

The proof for when $f$ has a maximum is given on page 224 of the textbook.

Q.E.D.

The maximum and minimum values of a function correspond to its optimum values. For instance, if ${\displaystyle C\left(x\right)}$ is a cost function of certain commodity $x$, then the local maximum and minimum values of ${\displaystyle C}$ on an interval $J$ correspond to the maximum and minimum costs when $x$ is limited to the interval $J$. The highest and lowest costs of ${\displaystyle C}$ correspond to the absolute maximum and minimum values of the function ${\displaystyle C}$ on the interval $J$.

Compare Definitions 5.7 and 5.9, and note that if a function has absolute extreme values on an interval, they are unique. Conversely, a function may have several local extreme values on an interval. A function may or may not have absolute extreme values on its domain.

Example 5.31. The function ${\displaystyle f\left(x\right)=x^2}$ has an absolute minimum at ${\displaystyle 0}$ (with an absolute minimum value of ${\displaystyle f\left(0\right)=0}$), but does not have an absolute maximum value on its domain.

Example 5.32. The function ${\displaystyle f\left(x\right)=x^3}$ does not have absolute extreme values on its domain. However, it does have them in the closed interval ${\displaystyle \left[-3\text{, }4\right]}$ because ${\displaystyle f\left(-3\right)\le f\left(x\right)}$ and ${\displaystyle f\left(4\right)\ge f\left(x\right)}$ for any $x$ in the interval ${\displaystyle \left(-3\text{, }4\right)}$.

Observe that the absolute extreme values ${\displaystyle f\left(-3\right)=-27}$ and ${\displaystyle f\left(4\right)=64}$ are attained at numbers not in the open interval ${\displaystyle \left(-3,4\right)}$. [Remember that the end points of an open interval are not included in the interval.]

Example 5.33. Although the function ${\displaystyle f\left(x\right)=1∕x}$ is positive on the interval ${\displaystyle \left[0,\infty \right)}$—that is ${\displaystyle f\left(x\right)>0}$ for all $x$ in ${\displaystyle \left(0,\infty \right)}$—it does not have an absolute minimum value, because there is no $c$ such that ${\displaystyle f\left(c\right)\le f\left(x\right)}$ for all $x$ in ${\displaystyle \left(0,\infty \right)}$.

Example 5.34. The absolute maximum value of the cosine function in its domain is ${\displaystyle 1}$, and is attained at infinitely many values $2k\pi \text{,}$ for any integer $k$.

However, in the interval ${\displaystyle \left[{\displaystyle \frac{\pi }{3}},\pi \right]}$ the absolute maximum value is ${\displaystyle \text{cos }\left({\displaystyle \frac{\pi }{3}}\right)={\displaystyle \frac{1}{2}}\text{.}}$

Example 5.35. A function may have local extreme values, but no absolute extreme ones. In the graph shown in Figure 5.12, there is one local maximum and one local minimum, but no absolute extreme values on the domain.

Figure 5.12. Function with local, but not absolute, extreme values on its domain

The following theorem gives conditions for a function to have absolute extreme values.

Theorem 5.10. The Extreme Value Theorem

A continuous function $f$ on a closed interval ${\displaystyle \left[a,b\right]}$ has absolute extreme values in the interval; that is, there are numbers $c$ and $d$ in the interval ${\displaystyle \left[a,b\right]}$ such that

${\displaystyle f\left(c\right)\ge f\left(x\right)\text{and}f\left(d\right)\le f\left(x\right)\text{ for all }x\text{ in }\left[a,b\right]\text{.}}$

The Extreme Value Theorem assure us that a continuous function on a closed interval has absolute extreme values, and from Figure 5 on page 223 of the textbook, we know that we should look for them at the endpoints, or at the points where the function has critical points. So, to find them, we use the closed interval method, described below.

Example 5.36. The function

${\displaystyle f\left(x\right)={\displaystyle \frac{x}{x^2-4}}}$

is continuous on ${\displaystyle \left[3,6\right]}$.

By the closed interval method

Example 5.37. The function ${\displaystyle g\left(x\right)=x^3-6x^2+9x+2}$ is continuous on the interval ${\displaystyle \left[-1,2\right]}$.

1. ${\displaystyle g^{\prime }\left(x\right)=3x^2-12x+9=3\left(x-3\right)\left(x-1\right)=0}$ if ${\displaystyle x=3}$ and ${\displaystyle x=1}$.
2. The only critical point in the interval ${\displaystyle \left[-1,2\right]}$ is ${\displaystyle x=1}$, thus ${\displaystyle g\left(1\right)=6}$.
3. ${\displaystyle g\left(-1\right)=-14}$ and ${\displaystyle g\left(2\right)=4}$.
4. The absolute maximum value is ${\displaystyle 6}$ (at ${\displaystyle x=1}$), and the absolute minimum value is ${\displaystyle -14}$ (at ${\displaystyle x=-1}$).

Footnote

^[1] For a discussion of the meaning of ${\displaystyle \delta }$ in this context, read through the section “The Definition of a Limit” in Unit 3.