Suppose that $f(x)$ is a continuous function over a closed interval [a, b].
To find the absolute maximum and minimum values of the function $f(x)$ on [a, b]:
(that is, find all x values for which $f'(x)=0$ (critical points), $f'(x)$ does not exist (possible cusps, etc.).
Find the absolute minimum and maximum values of the function, if they exist, over the given interval.
$f(x)={{x}^{2}}-6x-3\quad \quad [-1,5]$
$ f'(x)=2x-6 $
Criticals: $f'(x)=0 $
$ 2x-6=0 $
$ 2x=6 $
$ x=3 $
Calculate $f(x)$ for the interval endpoints: ($x=-1, 5$), and critical values: ($x=3$).
| $x$ | $f(x)$ |
|---|---|
| -1 | $ 4$ |
| 3 | $ -12 $ |
| $5$ | $ -8 $ |
The Absolute Maximum on the interval is: $(-1, 4)$.
The Absolute Minimum on the interval is: $(3, -12)$.
Find the absolute minimum and maximum values of the function, if they exist, over the given interval.
$f(x)=2{{x}^{3}}-3{{x}^{2}}-36x+62\quad \quad [-3,4]$
$ f'(x)=6{{x}^{2}}-6x-36 $
$ 6{{x}^{2}}-6x-36=0 $
$ 6\left( {{x}^{2}}-x-6 \right)=0 $
$ 6(x+2)(x-3)=0 $
$ x=-2$, and $x=3$
Calculate $f(x)$ for the interval endpoints: ($x=-3$ and $x= 4$), and critical values: ($x=-2$ and $x=3$).
| $x$ | $f(x)$ |
|---|---|
| -3 |
$ 89$ |
| -2 |
$106 $ |
| 3 |
$-19 $ |
| 4 |
$-2$ |
The Absolute Maximum on the interval is: $(-2, 106)$.
The Absolute Minimum on the interval is: $(3, -19)$.
Find the absolute minimum and maximum values of the function, if they exist, over the given interval.
$f(x)=x^3-9x^2+24x-23\quad \quad [-2,3]$
$ f(x)=x^3-9x^2+24x-23 $
$ f'(x)=3x^2-18x+24 $
$ f'(x)=3(x-4)(x-2) $
$ 3(x-4)(x-2)=0 $
$ x=2\text{ and }x=4 $
Calculate $f(x)$ for the interval endpoints: ($x=-2$ and $x=3$), and critical value(s): ($x=2$).
Do not include the critical value $x=4$ because it is not in the interval.
| $x$ | $f(x)$ |
|---|---|
| $-2$ |
$-115$ |
| $2$ |
$-3$ |
| $3$ |
$-5$ |
The Absolute Maximum on the interval is: $(2, -3)$.
The Absolute Minimum on the interval is: $(-2, -115)$.
$f(x)=x+\frac{1}{x}\quad \quad [1,20]$
$ f(x)=x+{{x}^{-1}} $
$ f'(x)=1+(-1){{x}^{-2}} $
$ f'(x)=1-\frac{1}{{{x}^{2}}} $
$ 0=1-\frac{1}{{{x}^{2}}} $
$ \frac{1}{{{x}^{2}}}=1 $
$ {{x}^{2}}=1 $
$ \sqrt{{{x}^{2}}}=1 $
$ x=1\text{ and }x=-1 $
Calculate $f(x)$ for the interval endpoints: ($x=1$ and $x=-1$), and critical values: ($x=20$).
Do not include the critical value $x=-1$ because it is not in interval.
| $x$ | $f(x)$ |
|---|---|
| $1$ |
$2$ |
| $20$ |
$20.05$ |
The Absolute Maximum on the interval is: $(20, 20.05)$.
The Absolute Minimum on the interval is: $(1, 2)$.
$f(x)=-3\quad \quad [-2,2]$
$f'(x)= 0$
| $x$ | $f(x)$ |
|---|---|
| $-2$ |
$-3$ |
| $2$ |
$-3$ |
There are no maximum or minimum values because this is a constant function.
An employee’s monthly production $M$, in number of units produced, is found to be a function of the number of year of service, $t$. For a certain product, a productivity function is given by: $M(t)=-2{{t}^{2}}+100t+180,\quad 0\le t\le 40$
Find the maximum productivity and the year in which it is achieved.
$M'(t)=-4t+100$
$ -4t+100=0$
$t=25$
Calculate $M(t)$ for the interval endpoints: ($t=0, 40$), and critical values: ($t=25$).
| $t$ | $M(t)$ |
|---|---|
| 0 |
$180$ |
| 25 |
$1430$ |
| 40 |
$980$ |
The maximum monthly productivity is 1430 units per month in year 25 of service.
Norris, Ken. (1999). Optimization Problems. Retrieved from https://www.stf.sk.ca/portal.jsp?Sy3uQUnbK9L2RmSZs02CjV/Lfyjbyjsxsd+sU7CJwaIY=F
$f(x)={{x}^{3}}-3x+6\quad \quad [-2,2]$
$f'(x)=3{{x}^{2}}-3$
$ 3{{x}^{2}}-3=0$
$ {{x}^{2}}=1$
$ x=\pm 1$
| $x$ | $f(x)$ |
|---|---|
| $-2$ | $4$ |
| $-1$ | $8$ |
| $1$ | $4$ |
| $2$ | $8$ |
In the interval, the function has an Absolute Minimum at the point: $(-2, 4)$ and $(1, 4).$
In the interval, the function has an Absolute Maximum at the point: $(-1, 8)$ abd $(2, 8).$
$y={{x}^{4}}-4{{x}^{3}}\quad \quad [-4,6]$
$y'=4{{x}^{3}}-12{{x}^{2}}$
$ 4{{x}^{3}}-12{{x}^{2}}=0$
$ 4{{x}^{2}}(x-3)=0$
$ 4{{x}^{2}}=0\quad x-3=0$
$ x=0\quad \quad x=3$
| $x$ | $y$ |
|---|---|
| $-4$ | $512$ |
| $0$ | $0$ |
| $3$ | $-27$ |
| $6$ | $432$ |
In the interval, the function has an Absolute Minimum at the point: $(3, -27).$
In the interval, the function has an Absolute Maximum at the point: $(-4, 512).$
Find the absolute minimum and maximum values of the function, if they exist, over the given interval.
$g'(x)=x^2-2$
$ {{x}^{2}}-2=0 $
$ x=\sqrt{2},x=-\sqrt{2} $
| $x$ | $g(x)$ |
|---|---|
| $-2$ | $-.67$ |
| $-1.414$ | $-.11$ |
| $1.414$ | $-3.89$ |
| $3$ | $1$ |
In the interval, the function has an Absolute Minimum at the point: $(1.414,-3.89).$
In the interval, the function has an Absolute Maximum at the point: $(3, 1).$
| $x$ | $g(x)$ |
|---|---|
| $-2$ | $-.67$ |
| $-1.414$ | $-.11$ |
| $1.414$ | $-3.89$ |
| $2$ | $-3.33$ |
In the interval, the function still has an Absolute Minimum at the point: $(1.414, -3.89).$
In the interval, the function now has an Absolute Maximum at the point: $(-1.414, -.11).$
| $x$ | $g(x)$ |
|---|---|
| $-2$ | $-.67$ |
| $-1.414$ | $-.11$ |
| $1$ | $-3.67$ |
In the interval, the function now has an Absolute Minimum at the point: $(1, -3.67).$
In the interval, the function now has an Absolute Maximum at the point: $(-1.414, -.11).$
The temperature, T, of person during an illness is given by: $$T(t)=-0.1{{t}^{2}}+1.2t+98.6\ \ \ 0\le t\le 12$$ where T = temperature (°F) at time t, in hours since noon. Find the maximum value of the temperature and when it occurs.
$ T'(t)=-0.2t+1.2 $
$ -0.2t+1.2=0 $
$ -0.2t=-1.2 $
$ t=6 $
| $t$ | $T(t)$ |
|---|---|
| $0$ | $98.6$ |
| $6$ | $102.2$ |
| $12$ | $98.6$ |
In the interval, the maximum temperature of 102.2 degrees occurs at 6 pm.