For what values of x is #f(x)=(-2x^2)/(x-1)# concave or convex?

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Answer 1 · 2018-07-27T21:17:26

The function will be concave in #(1, \infty)#

The function will be convex in #(- \infty, 1)#

Given function:

#f(x)=\frac{-2x^2}{x-1}#

#f'(x)=\frac{(x-1)(-4x)-(-2x^2)(1)}{(x-1)^2}#

#=\frac{-2x^2+4x}{(x-1)^2}#

#f''(x)=\frac{(x-1)^2(-4x+4)-(-2x^2+4x)(2(x-1))}{(x-1)^4}#

#f''(x)=-4/(x-1)^3#

The function will be concave iff #f''(x)<0#

#\therefore -4/(x-1)^3< 0#

#1/(x-1)^3> 0#

#x\in (1, \infty)#

The function will be concave in #(1, \infty)#

The function will be convex in #(- \infty, 1)#