Math 22 Extrema and First Derivative Test

Relative Extrema

 Let  $f$  be a function defined at  $c$ .
 1.  $f(c)$  is a relative maximum of  $f$  when there exists an interval  $(a,b)$  containing  $c$  such that  $f(x)\leq f(c)$  for all  $x$  in  $(a,b)$ .
 2.  $f(c)$  is a relative minimum of  $f$  when there exists an interval  $(a,b)$  containing  $c$  such that  $f(x)\geq f(c)$  for all  $x$  in  $(a,b)$ .

If $f$ has a relative minimum or relative maximum at $x=c$ , then $c$ is a critical number of $f$ . That is, either $f'(c)=0$ or $f'(c)$ is undefined.

Relative extrema must occur at critical numbers as shown in picture below.

The First-Derivative Test

 Let  $f$  be continuous on the interval  $(a,b)$  in which  $c$  is the only critical number, then
 
 On the interval  $(a,b)$ , if  $f'(x)$  is negative to the left of  $x=c$  and positive to the right of  $x=c$ , then  $f(c)$  is a relative minimum.
 
 On the interval  $(a,b)$ , if  $f'(x)$  is positive to the left of  $x=c$  and negative to the right of  $x=c$ , then  $f(c)$  is a relative maximum.

Guidelines for Finding Relative Extrema

 1. Find the derivative of  $f$ 
 2. Find all critical numbers, then determine the test intervals
 3. Determine the sign of  $f'(x)$  at an arbitrary number in each test intervals
 4. Apply the first derivative test

Exercises: Find all relative extrema of the functions below

1) $f(x)=x^{2}+2x$

Solution:
Step 1: $f'(x)=2x+2=2(x+1)=0$ ,
Step 2: Critical number is $x=-1$ , so the test intervals are $(-\infty ,-1)$ and $(-1,\infty )$
Step 3: Choose $x=-2$ for the interval $(-\infty ,-1)$ , and $x=0$ for the interval $(-1,\infty )$ .
Then we have: $f'(-2)=-2<0$ and $f'(0)=2>0$
Step 4: By the first derivative test, $f'(x)$ is negative to the left of $x=-1$ and positive to the right of $x=-1$ , then $f(1)=3$ is a relative minimum
Therefore, Relative minimum: $(1,3)$
(Note: $f(x)$ in this case is a parabola so our answer makes sense)

2) $f(x)=5x^{3}-10x^{2}+3$

Solution:
Step 1: $f'(x)=15x^{2}-20x=5x(3x-4)=0$ ,
Step 2: Critical number is $x=0$ and $x={\frac {4}{3}}$ , so the test intervals are $(-\infty ,0),(0,{\frac {4}{3}})$ and $({\frac {4}{3}},\infty )$
Step 3: Choose $x=-1$ for the interval $(-\infty ,0)$ , $x=1$ for the interval $(0,{\frac {4}{3}})$ and $x=2$ for the interval $({\frac {4}{3}},\infty )$ .
Then we have: $f'(-1)=35>0$ , $f'(1)=-5<0$ and $f'(2)=20>0$
Step 4: By the first derivative test, $f'(x)$ is positive to the left of $x=0$ and negative to the right of $x=0$ , then $f(0)=3$ is a relative maximum,
and $f'(x)$ is negative to the left of $x={\frac {4}{3}}$ and positive to the right of $x={\frac {4}{3}}$ , then $f({\frac {4}{3}})={\frac {-79}{27}}$ is a relative minimum.
Therefore, Relative minimum: $({\frac {4}{3}},{\frac {-79}{27}})$ and Relative maximum: $(0,3)$

Absolute Extrema

 Let  $f$  be defined on an interval  $I$  containing  $c$ .
 1.  $f(c)$  is an absolute minimum of  $f$  on  $I$  when  $f(c)\leq f(x)$  for every  $x$  in  $I$ 
 2.  $f(c)$  is an absolute maximum of  $f$  on  $I$  when  $f(c)\geq f(x)$  for every  $x$  in  $I$

Extreme Value Theorem

 If  $f$  is continuous on a closed interval  $[a,b]$ , then  $f$  has both a minimum value and a maximum value on  $[a,b]$ .

Guidelines for Finding Extrema on a Closed Interval

 To find the extrema of a continuous function  $f$  on a closed interval  $[a,b]$ , use the following steps.
 1. Find all critical numbers of  $f$ 
 2. Evaluate  $f$  at each of its critical number
 3. Evaluate  $f$  at each end point  $a$  and  $b$ 
 4. The least of these values is the absolute minimum, and the greatest is the maximum.

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