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

2 Answers
Jul 7, 2018

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.

Jul 7, 2018

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]}