Question

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

Answer 1

local min is at `(0, 0)`

There is no local maximum.

Explanation:

Given: `f(x) = x^2/(x-2)^2`

Find the first derivative:

Use the quotient rule of differentiation: `(u/v)' = (vu' - u v')/v^2`

Let `u = x^2; " "u' = 2x; " "v = (x-2)^2; " "v' = 2(x-2)^1(1)`

`f'(x) = ((x-2)^2(2x) - 2x^2(x-2))/(x-2)^4`

`f'(x) = ((x-2)[2x(x-2) - 2x^2])/(x-2)^4`

`f'(x) = (2x^2 - 4x - 2x^2)/(x-2)^3 = (-4x)/(x-2)^3`

Find the critical point(s). Let `f'(x) = 0`:

`f'(x) = (-4x)/(x-2)^3 = 0`

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

critical value: `x = 0`

`f(0) = 0`

critical point: `(0, 0)`

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

intervals: `" "(-oo, 0), " "x = 0, " "(0, oo)`
test value: `" " x = -1 " "x = 0 " "x = 1`
`f'(x): " " < 0 " "= 0" " > 0`

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

local min is at `(0, 0)`

There is no local maximum.

Answer 2

There is a local min at `(0,0)`

Explanation:

The function is

`f(x)=(x^2)/(x-2)^2`

This function is a quotient of `2` derivable functions

The derivative of a quotient is

`(u/v)'=(u'v-uv')/(v^2)`

`u=x^2`, `=>`, `u'=2x`

`v=(x-2)^2`, `=>`, `u'=2(x-2)`

Therefore,

`f'(x)=((2x)(x-2)^2-(x-2)(2x^2))/(x-2)^4`

`=(2x(x-2)-2x^2)/(x-2)^3`

`=(2x^2-4x-2x^2)/(x-2)^3`

`=-(4x)/(x-2)^3`

The critical points are when

`f'(x)=0`

That is

`x=0`

Let's build a variation chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaa)` `0` `color(white)(aaaaaaaa)` `2` `color(white)(aaaaa)` `+oo`

`color(white)(aaaa)` `f'(x)` `color(white)(aaaa)` `-` `color(white)(aaa)` `0` `color(white)(aaa)` `+` `color(white)(aa)` `||` `color(white)(aaa)` `-`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaa)` `↘` `color(white)(aa)` ###color(white)(aaa)#`↗` `color(white)(aaa)` `||` `color(white)(aa)` `↘`

There is a local min at `(0,0)`

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