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.

Exercises: Find all absolute extrema of the function below

1) $f(x)=5-2x^{2}$ on $[-3,1]$

Solution:
Step 1: $f'(x)=-4x=0$, So, critical number is $x=0$
Step 2: $f(0)=5$
Step 3: $f(-3)=-13$ and $f(1)=3$
Step 4: Absolute Maximum is $5$ at $x=0$
and absolute minimum is $-13$ at $x=-3$

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