# How do you find the local max and min for f (x) = x^(3) - 6x^(2) + 5?

Dec 18, 2015



#### Explanation:

Find the critical values (when $f ' \left(x\right) = 0$ or is undefined).

$f ' \left(x\right) = 3 {x}^{2} - 12 x$

$f ' \left(x\right) = 3 x \left(x - 4\right)$

$f ' \left(x\right) = 0$ when $x = 0 , 4$.

Make a sign chart for $f ' \left(x\right)$.

$\textcolor{w h i t e}{\times \times \times \times \times \times} 0 \textcolor{w h i t e}{\times \times \times \times \times \times} 4$
$\leftarrow - - - - - - - - - - - - - - - - \rightarrow$
$\textcolor{w h i t e}{\times x} \text{POSITIVE"color(white)(xxxxx)"NEGATIVE"color(white)(xxxxx)"POSITIVE}$

There is a relative maximum at $x = 0$ because the derivative goes from positive to negative.

There is a relative minimum at $x = 4$ because the derivative goes from negative to positive.

Look at a graph:

graph{x^3-6x^2+5 [-36.65, 55.83, -32.08, 14.18]}

Dec 18, 2015

It has a local maxima at $\left(0 , 5\right)$
It has a local minima at $\left(4 , - 27\right)$

#### Explanation:

Given -

$y = {x}^{3} - 6 {x}^{2} + 5$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} - 12 x$
$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 6 x - 12$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies 3 {x}^{2} - 12 x = 0$
$3 x \left(x - 4\right) = 0$

$3 x = 0$
$x = 0$

$x - 4 = 0$
$x = 4$

AT x=0; (d^2y)/dx^2=6(0)-12=-12 <0
At x=0; dy/dx=0;(d^2)/dx^2<0 Hence the function has a local maximum at $x = 0$

AT x=4;(d^2y)/dx^2=6(4)-12=24-12=12>0

Local Maximum is -
At x=0; y=(0)^3-6(0)^2+5=5

$\left(0 , 5\right)$

AT x=4; dy/dx=0;(d^2y)/dx^2>0. Hence the function has a local minimum.

Local Minimum -
At x=4;y=(4)^3-6(4)^2+5=64-96+5=-27#
$\left(4 , - 27\right)$

graph{x^3-6x^2+5 [-58.5, 58.55, -29.24, 29.3]}