Using derivatives

The derivative tells you what a function is doing: rising, falling, or momentarily flat. At a peak or valley, the slope is zero, which is why derivatives solve optimization problems.

Sign of $f^{'}$	Function
Positive	Rising
Negative	Falling
Zero, + to −	Local maximum
Zero, − to +	Local minimum

Work through it

Find local extremes of $f (x) = x^{3} - 3 x$ .

Derivative. $3 x^{2} - 3$

Set to zero. $x = \pm 1$

Test $f^{'}$ between them: $f^{'} (0) = - 3$ .falling on $(- 1, 1)$

So.max at $x = - 1$ , min at $x = 1$

A farmer has 40 m of fence for a rectangle. Largest area?

Express area with one variable. $A = x (20 - x)$

Differentiate. $A^{'} = 20 - 2 x$

Set to zero. $x = 10$

Area. $100$ m², a square

Position $s = t^{3} - 6 t^{2}$ . When is velocity zero?

Velocity is the derivative. $v = 3 t^{2} - 12 t$

Solve. $t = 0$ or $t = 4$

Watch out for this

Assuming every $f^{'} = 0$ is a max or min.

$f (x) = x^{3}$ has $f^{'} (0) = 0$ , but it keeps rising. Check the sign on each side.

The short version

f^{'} > 0

means increasing and

f^{'} < 0

means decreasing. Maximums and minimums happen where

f^{'} = 0

, or at the endpoints.

Your turn

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