For what values of x is #f(x)= (x-x^3)/(2-x^3)# concave or convex?

1 Answer
bp
Aug 1, 2016

f(x) is concave down in (0,#2^(1/3)#) and concave up in (#2^(1/3), oo#)
&
concave up in (0, #-oo#)

Explanation:

f'' (x) would be #-6x((x^4 -4x^3 +4x -4))/(x^3-2)^3#

For concavity, the second derivative test says that for f(x) is concave up for that value of x for which f''(x)>0 and concave down if f''(x)<0.
From the 2nd derivative shown above, it can be ascertained that for x>0 and up to x<#2^(1/3)#, f''(x) would be <0, at x=#2^(1/3)#, f(x) would not exist and for all x>#2^(1/3)# f''(x)>0.

Hence f(x) is concave down in (0,#2^(1/3)#) and concave up in (#2^(1/3), oo#)

Like wise for all x<0, f''(x) would be positive and hence concave up.

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