How do you find the local max and min for #f(x)=x+(1/x)#?

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Answer 1 · 2017-03-06T04:41:20

Local #f_min= (1, 2)#
Local #f_max = (-1, -2)#

#f(x) =x+1/x#

#f(x)# will have extrema where #f'(x)=0#

#f'(x) = 1-1/x^2# [Power rule]

#1-1/x^2 =0 -> x^2=1#

#x= +-1#

As can be seen by the graph of #f(x)# below, #f(x)# has a local maximum at #x=-1# and a local minimum at #x=+1#

graph{x+1/x [-12.66, 12.65, -6.33, 6.33]}

Therefore:
Local #f_min = f(1) = 1+1 =2 -> (1,2)#
and
Local #f_max = f(-1) = -1-1 =-2-> (-1,-2)#