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: $f'(x)$ is negative to the left of $x=-1$ and positive to the right of $x=-1$, then $f(1)=3$ is a relative minimum
(Note: $f(x)$ in this case is a parabola so our answer makes sense)

2) $f(x)=6x^{\frac {2}{3}}+4x^{2}+3$

Solution:
$f(x)=x^{\frac {1}{2}}$
So, $f'(x)={\frac {1}{2}}x^{\frac {-1}{2}}={\frac {1}{2{\sqrt {x}}}}$
In this case, we have critical number when $f'(x)$ is undefined, which is when ${\sqrt {x}}=0$. So critical number is $x=0$

Absolute Extrema

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