Critical number, maximum and minimum values

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How to describe graphs of functions?
$\bullet$ local maximum
$\bullet$ local minimum
$\bullet$ critical number
state the:
$\bullet$ absolute maximum
$\bullet$ absolute minimum
on the interval,
$-1\leq x\leq 12\,$ state the:
$\bullet$ absolute maximum
$\bullet$ absolute minimum

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Find the critical numbers of the function:
1. $f(x)={^3}\sqrt{x^{2}-5x}$
2. $f(x)=x^{\frac{1}{3}}-x^{-\frac{2}{3}}$

First Derivative Test: a test to determine whether or not

\ f

has a local maximum or minimum at a critical number

First Derivative Test
local maximum	local minimum	no maximum or minimum	no maximum or minimum

$f(x)=3x^{5}-15x^{4}+25x^{3}-15x^{2}+5$ $f (x) = 3 x^{5} - 15 x^{4} + 25 x^{3} - 15 x^{2} + 5$
1. Find the critical numbers.
2. On what intervals is $f$ increasing or decreasing?
3. Find the local maximum and minimum values.
4. Sketch the graph.
5. Find the absolute maximum and minimum values.
The Closed Interval Method
To find the absolute maximum and minimum values of a continuous function $f$ $f$ on a closed interval [a, b]:
1.Find the values of $f$ $f$ at the critical numbers of $f$ $f$ in (a, b).
2.Find the values of $f$ $f$ at the left-endpoint and right-endpoint of the interval
3.Compare all values from steps 1 and 2: the largest $\$ is the absolute maximum value; the smallest $\$ is the absolute minimum value.
Find the absolute maximum and minimum values of the function:

$f(x)=3x^{5}-15x^{4}+25x^3-15x^2+5$ $f (x) = 3 x^{5} - 15 x^{4} + 25 x^{3} - 15 x^{2} + 5$
$-\frac{1}{2}\leq x\leq\frac{1}{2}$ $- \frac{1}{2} \leq x \leq \frac{1}{2}$