What are the local extrema of #f(x)= x/((x-2)(x-4)^3)#?

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Answer 1 · 2016-01-17T07:49:11

#x_1=2.430500874043# and #y_1=-1.4602879768904# Maximum Point

#x_2=-1.0971675407097# and #y_2=-0.002674986072485# Minimum Point

Determine the derivative of #f(x)#

#f' (x)#
#=((x-2)(x-4)^3*1-x[(x-2)*3(x-4)^2 +(x-4)^3 *1] )/[(x-2)(x-4)^3]^2#

Take the numerator then equate to zero

#((x-2)(x-4)^3*1-x[(x-2)*3(x-4)^2 +(x-4)^3 *1] )=0#

simplify

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

Factoring the common term

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

#(x-4)^2*(x^2-6x+8-3x^2+6x-x^2+4x)=0#

#(x-4)^2(-3x^2+4x+8)=0#

The values of x are:

#x=4# an asymptote

#x_1=(4+sqrt(112))/6=2.430500874043#
Use #x_1# to obtain #y_1=-1.4602879768904# Maximum

#x_2=(4-sqrt(112))/6=-1.0971675407097#
Use #x_2# to obtain #y_2=-0.002674986072485## Minimum