What are the local extrema of #f(x)= x^2/(x^2-3x-5) #?

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What are the local extrema of #f(x)= x^2/(x^2-3x-5) #?

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Answer 1 · 2018-05-27T17:17:25

#MAX(0;0)# and #MIN(-10/3,20/29)#

Explanation:

We compute
#f'(x)=-x(3x+10)/(x^2-3x-5)^2#
#f''(x)=2(3x^2+15x^2+25)/(x^2-3x-5)^3#
so
#f'(x)=0#
if
#x=0# or
#x=-10/3#
we have further
#f''(0)=-2/5<0#
and
#f''(-10/3)=162/4205>0#

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What are the local extrema of #f(x)= x^2/(x^2-3x-5) #?

1 Answer

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