How do you find the local extrema for #y = [1 / x] - [1 / (x - 1)]#?

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Answer 1 · 2017-03-08T12:41:56

#y# has a local minimum of 4 at #x=1/2#

#y=1/x - 1/(x-1)#

#= x^(-1) - (x-1)^(-1)#

#y' = -x^-2 + (x-1)^-2 *1# {Power rule and Chain rule]

#y# will have extrema where: #y' =0#

I.e. where: #-x^-2 + (x-1)^-2 =0#

#(-(x-1)^2+x^2)/(x^2(x-1)^2)=0#

#-x^2+2x-1 + x^2 =0#

#x=1/2#

As can be seen from the graph of #y# below, #x=1/2# is local minimum.

graph{1/x - 1/(x-1) [-10.66, 11.84, -3.845, 7.405]}

Thus: #y_min = y(1/2) = 1/(1/2) - 1/(1/2 -1)#

#= 2+2 =4#