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

Question

1675 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-04T08:46:20

The function #f(x)# is convex at #x=-1#

The derivative of a quotient is

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

We calcule the first derivative

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

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

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

Now, we calculate the second derivative,

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

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

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

At #x=-1#

#f''(-1)=e^(-1-1)-(2)/(-1-1)^3=0.135+0.25=0.385#

As #f''(-1)>0#, we conclude that #f(x)# is convex at #x=-1#

graph{e^(x-1)-x^2/(x-1)-1 [-10, 10, -5, 5]}