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

##### 2 Answers

local min is at

There is no local maximum.

#### Explanation:

Given:

**Find the first derivative:**

Use the quotient rule of differentiation:

Let

Find the critical point(s). Let

#-4x = 0 * (x-2)^3 = 0$

critical value:

**critical point:**

Use the first derivative test. Set up intervals and test values in the intervals to see if they are positive or negative:

intervals:

test value:

When the slope changes from negative, to zero, to positive, we have a local minimum.

local min is at

There is no local maximum.

There is a local min at

#### Explanation:

The function is

This function is a quotient of

The derivative of a quotient is

Therefore,

The critical points are when

That is

Let's build a variation chart

There is a local min at

graph{x^2/(x-2)^2 [-11.82, 13.49, -3.49, 9.17]}