Is #f(x)=e^x-x^2/(x-1)# concave or convex at #x=0#?

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Answer 1 · 2018-07-09T14:42:28

The function is convex at #x=0#

The derivative of a quotient is

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

The function is

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

The first derivative is

#f'(x)=e^x-(2x(x-1)-x^2*1)/(x-1)^2#

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

The second derivative is

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

#=e^x-((2x-2)(x-1)-(2x^2-4x))/(x-1)^3#

#=e^x-(2x^2-4x+2-2x^2+4x)/(x-1)^3#

#=e^x-2/(x-1)^3#

When #x=0#

#f''(0)=e^0-2/(-1)^3=1-2=3#

As #f''(0) >0#, the function is convex at #x=0#

graph{e^x-x^2/(x-1) [-14.24, 14.23, -7.12, 7.12]}