# Math 22 Extrema and First Derivative Test

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## Relative Extrema

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


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

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

## The First-Derivative Test

 Let ${\displaystyle f}$ be continuous on the interval ${\displaystyle (a,b)}$ in which ${\displaystyle c}$ is the only critical number, then

On the interval ${\displaystyle (a,b)}$, if ${\displaystyle f'(x)}$ is negative to the left of ${\displaystyle x=c}$ and positive to the right of ${\displaystyle x=c}$, then ${\displaystyle f(c)}$ is a relative minimum.

On the interval ${\displaystyle (a,b)}$, if ${\displaystyle f'(x)}$ is positive to the left of ${\displaystyle x=c}$ and negative to the right of ${\displaystyle x=c}$, then ${\displaystyle f(c)}$ is a relative maximum.


## Guidelines for Finding Relative Extrema

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


Exercises: Find all relative extrema of the functions below

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

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

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

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

## Absolute Extrema

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


## Extreme Value Theorem

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


## Guidelines for Finding Extrema on a Closed Interval

 To find the extrema of a continuous function ${\displaystyle f}$ on a closed interval ${\displaystyle [a,b]}$, use the following steps.
1. Find all critical numbers of ${\displaystyle f}$
2. Evaluate ${\displaystyle f}$ at each of its critical number
3. Evaluate ${\displaystyle f}$ at each end point ${\displaystyle a}$ and ${\displaystyle b}$
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) ${\displaystyle f(x)=5-2x^{2}}$ on ${\displaystyle [-3,1]}$

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