What are the local maxima and minima of #f(x)= (x^2)/(x-2)^2#?

#f(x)=x^2/{(x-2)^2#

This function has a vertical asymptote at #x=2#, because the denominator is zero when #x=2#.

It approaches #1# from above as x goes to #+oo # (horizontal asymptote) and approaches #1# from below as x goes to #-oo #, because for large values #x^2~=(x-2)^2# with #x^2>(x-2)^2# for #x>0# and #x^2<(x-2)^2# for #x<0#.

#{d f(x)}/dx =d/dx ( x^2/{(x-2)^2})# Use the quotient rule! #{d f(x)}/dx = ( {(d/dx x^2) (x-2)^2 - x^2 ( d/dx (x-2)^2) }/{(x-2)^4})#. Using rule for powers and the chain rule we get: #{d f(x)}/dx = {(2x) (x-2)^2 - x^2 (2*(x-2)*1) }/(x-2)^4#. We now neaten up a bit ... #{d f(x)}/dx = {2x(x^2-4x+4) - x^2(2x-4) }/(x-2)^4# #{d f(x)}/dx = {2x^3-8x^2+8x - 2x^3+4x^2 }/(x-2)^4# #{d f(x)}/dx = {-4x^2 + 8x }/(x-2)^4#

Now the second derivative, done like the first. #{d^2 f(x)}/dx^2 = {d/dx(-4x^2 + 8x)(x-2)^4 - (-4x^2 + 8x)(d/dx ((x-2)^4))}/(x-2)^8# #{d^2 f(x)}/dx^2 = {(-8x + 8)(x-2)^4 - (-4x^2 + 8x)(4(x-2)^3 *1)}/(x-2)^8# #{d^2 f(x)}/dx^2 = {(-8x + 8)(x-2)^4 - (-4x^2 + 8x)(4(x-2)^3 *1)}/(x-2)^8# It's ugly but we only need to plug and and note where it's badly behaved.

#{d f(x)}/dx = {-4x^2 + 8x }/(x-2)^4# This function is undefined at #x=2#, that asymptote, but looks fine everywhere else. We want to know were the max/min are ... we set #{d f(x)}/dx=0# #{-4x^2 + 8x }/(x-2)^4=0# this is zero when the numerator is zero and if the denominator is not. #-4x^2 + 8x=0# #4x( -x+2)=0# or #4x(2-x)=0# This is zero at #x=0# and #x=2#, but we cannot have a max/min were the derivative/function are undefined, so the only possibility is #x=0#.

"the second derivative test" Now we look at the second derivative, ugly as it is ... #{d^2 f(x)}/dx^2 = {(-8x + 8)(x-2)^4 - (-4x^2 + 8x)(4(x-2)^3)}/(x-2)^8# Like the function and the first derivative this is undefined at #x=2#, but looks fine everywhere else. We plug #x=0# into #{d^2 f(x)}/dx^2# #{d^2 f(0)}/dx^2=# # {(-8*0 + 8)(0-2)^4 - (-4*0^2 + 8*0)(4*0-2)^3}/(0-2)^8 # #= {(8)(-2)^4}/(2)^8 #, isn't zero such a lovely number to plug it? #=128/256# all that for #1/2#

#1/2 >0# so #x=0# is a local minima. To find the y value we need to plug it into the function. #f(x)=0^2/{(0-2)^2}=0# The origin!

To find the local extrema of \( f(x) = \frac{x^2}{(x-2)^2} \), we first find its critical points by setting the derivative equal to zero and solving for \( x \). Then, we check the nature of these critical points to determine if they are local maxima or minima. \( f'(x) = \frac{d}{dx} \left( \frac{x^2}{(x-2)^2} \right) \) \( = \frac{(x-2)^2(2x) - x^2(2(x-2))}{(x-2)^4} \) \( = \frac{2x(x-2)(x-2-2x)}{(x-2)^4} \) \( = \frac{-2x(x-2)}{(x-2)^4} \) \( = \frac{-2x}{(x-2)^3} \) Setting \( f'(x) \) equal to zero and solving for \( x \): \( \frac{-2x}{(x-2)^3} = 0 \) \( \Rightarrow -2x = 0 \) \( \Rightarrow x = 0 \) The critical point is \( x = 0 \). Now, let's analyze the nature of this critical point using the first derivative test. We'll examine the sign of \( f'(x) \) around \( x = 0 \). For \( x < 0 \), \( f'(x) < 0 \) (negative). For \( 0 < x < 2 \), \( f'(x) > 0 \) (positive). For \( x > 2 \), \( f'(x) < 0 \) (negative). This indicates that \( f(x) \) has a local maximum at \( x = 0 \) and a local minimum at \( x = 2 \).