Is #f(x)=(x^2-2)/(x-1)# increasing or decreasing at #x=2#?

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Answer 1 · 2017-06-09T19:10:08

The function is increasing

We calculate the first derivative of #f(x)#, plug in the value #x=2# and look at the sign.

#f(x)=(x^2-2)/(x-1)#

#u=(x^2-2)#, #=>#, #u'=2x#

#v=(x-1)#, #=>#, #v'=1#

#f'(x)=(u'v-uv')/(v^2)#

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

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

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

Therefore,

#f'(2)=(4-4+2)/(1)=2#

As #f'(2)>0#, so the function is increasing