# How do you find all critical point and determine the min, max and inflection given f(x)=x^3-6x^2+9x+8?

Apr 19, 2018

See below

#### Explanation:

Given $f \left(x\right)$, his critical point are given by f´(x)=0. Lets calculate

f´(x)=3x^2-12x+9=0

Using quadratic formula $x = \frac{12 \pm \sqrt{144 - 108}}{6} = \frac{12 \pm 6}{6}$

One critical point is $x = 3$ and the other is $x = 1$

Analyzing the sign of f´(x) in intervals $\left(- \infty , 1\right)$ $\left(1 , 3\right)$ and $\left(3 , + \infty\right)$ we found

f´(x)>0 in $\left(- \infty , 1\right)$ so f is increasing there
f´(x)<0 in $\left(1 , 3\right)$ so f is decreasing there
f´(x)>0 in $\left(3 , + \infty\right)$ so f is increasing there

Sumarizing $f \left(x\right)$ has a maximum in $x = 1$, has a minimum in $x = 3$ graph{x^3-6x^2+9x+8 [-15.25, 25.33, -3, 17.27]}

If we calculate f´´(x)=6x-12=0 we found x=2 as infexion point