Given function is f(x)=

= 1-

Hence f ’(x) =

And f “ (x) = -6

There are no points at which f” (x) =0, but at x= +√3 and x= -√3, f (x) is not continuous.

We can test for concavity in the intervals (-∞, -√3), (-√3, √3) and (√3, ∞) as shown in the table below.

Interval	-∞<x<-√3	-√3<x<√3	√3<x<∞
Test value	X=-2	X=0	X=2
Sign of f”(x)	-ive [f”(-2)<0]	+ive [f”(0)>0]	-ive [f”(2)<0]
Conclusion	Concave down	Concave up	Concave down

Hence point of inflexion are (√3,∞) and (-√3,∞)