How do use the first derivative test to determine the local extrema #y = (x^2 + 2) /( x^2 + 1)#?

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Answer 1 · 2015-10-21T07:31:44

#y_max=2#

#y' = ((x^2+2)' * (x^2+1)-(x^2+2) * (x^2+1)')/(x^2+1)^2#

#y' = (2x * (x^2+1)-(x^2+2) * 2x)/(x^2+1)^2#

#y' = (2x * (x^2+1-x^2-2))/(x^2+1)^2#

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

#y'=0 <=> -2x=0 <=> x=0#

#AAx>0 => y'<0#

#AAx<0 => y'>0#

The first derivative changes sign at #x=0# so it's the extrema point:
#y_max=y(0)=2/1=2#