# How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for f(x)=x^3-3x^2+1?

Dec 8, 2016

At $x = 0$ The function has a relative maximum.

At $x = 2$ the function has a relative minimum.

#### Explanation:

Given -

$y = {x}^{3} - 3 {x}^{2} + 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} - 6 x$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 6 x - 6$

At any given point, if $\frac{\mathrm{dy}}{\mathrm{dx}} > 0$ the function is increasing otherwise it is decreasing.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies 3 {x}^{2} - 6 x$
$3 x \left(x - 2\right) = 0$

$3 x = 0$
$x = 0$

$x - 2 = 0$
$x = 2$

At x=0; (d^2y)/(dx^2)=(6(0)-6=-6<0

At x=0; dy/dy =0;(d^2y)/(dx^2)<0

The function has a relative maximum.

At x=2; (d^2y)/(dx^2)=(6(2)-6=6>0

At x=2; dy/dy =0;(d^2y)/(dx^2)>0
There is a relative minimum.

graph{x^3-3x^2+1 [-10, 10, -5, 5]}