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

Question

Isabella Ackerman · Accepted Answer

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

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.

Emily Ammerman · Answer

undefined To determine the concavity or convexity of $ f(x) = \frac{x - x^3}{2 - x^3} $, we need to find its second derivative and then analyze its sign.

First, find the first derivative of $ f(x) $ and the second derivative by differentiating again. Then, set the second derivative equal to zero to find the inflection points.

The first derivative of $ f(x) $ is:
$$ f'(x) = \frac{(2 - x^3)(1 - 3x^2) - (x - x^3)(-3x^2)}{(2 - x^3)^2} $$

The second derivative of $ f(x) $ is:
$$ f''(x) = \frac{6x^4 - 4x^2 - 6x^2 + 2}{(2 - x^3)^3} $$

Now, set $ f''(x) = 0 $ and solve for $ x $ to find the inflection points.

$$ 6x^4 - 10x^2 + 2 = 0 $$

This is a quadratic equation in terms of $ x^2 $. Solve it to find $ x^2 $ and then find the corresponding values of $ x $.

Once you have the values of $ x $, analyze the sign of $ f''(x) $ in the intervals determined by the critical points and the boundary points (where the function is defined). If $ f''(x) > 0 $, the function is convex in that interval. If $ f''(x) < 0 $, the function is concave in that interval.