A function #f(x)#, whicg is twice continuously differentiable on an interval , then the function ݂is
- convex if #f''(x)>0# for all #x# in the interval, and
- concave if #f''(x)<0# for all #x# in the interval.
Now as #f(x)=(4x)/(x^2-1)#
#f'(x)=(4(x^2-1)-4x*2x)/(x^2-1)^2=-(1+4x^2)/(x^2-1)^2#
and #f''(x)=-(8x(x^2-1)^2-(1+4x^2)(4x(x^2-1)))/(x^2-1)^4#
= #-(8x^5-16x^3+8x-(16x^5+4x^3-16x^3-4x))/(x^2-1)^4#
= #-(-8x^5-4x^3+12x)/(x^2-1)^4=(4x(2x^4+x^2-3))/(x^2-1)^4#
= #(4x(2x^2+3)(x^2-1))/(x^2-1)^4=(4x(2x^2+3))/(x^2-1)^3#
Observe that #f''(x)<0# when #x in(-oo,-1)uu(0,1)# and hence in this region #f(x) is concave and
#f''(x)>0# when #x in(-1,0)uu(0,oo)# and hence in this region #f(x) is convex.
graph{(4x)/(x^2-1) [-10, 10, -5, 5]}
