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

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Answer 1 · 2017-11-27T14:20:09

concave

Let's calculate the second derivative and then find the sign of it when #x=-1#:

The first derivative is:

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

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

#=(x^2-2x-8)/(x-1)^2#

The second derivative is:

#f''(x)=((2x-2)(x-1)^2-(x^2-2x-8)*2(x-1))/(x-1)^4#

#=(2(x-1)^3-2(x-1)(x^2-2x-8))/(x-1)^4#

#=(2cancel((x-1))(cancelx^2cancel(-2x)+1cancel(-x^2)+cancel(2x)+8))/(x-1)^(cancel4^3)#

#=18/(x-1)^3#

that's less than 0 at #x=-1#:

#f''(-1)=18/(-1-1)^3=18/-8=-9/4<0#

Then the given function is concave at #x=-1#

graph{(x+2)^2/(x-1) [-10, 10, -20, 20]}