How do you find points of inflection and determine the intervals of concavity given #y=x^(1/3)(x-4)#?

Question

1660 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-08T18:30:15

There is a point of inflection at #x=2#. The graph is concave down for #x < 2#, and concave up for #x > 2#

Let's say #y = f(x) = x^{1/3}(x-4) = x^{4/3} - 4x^{1/3}#

This means:

#f'(x) = 4/3x^{1/3}-4/3x^{-2/3}#

and

#f''(x) = 4/9x^{-2/3}+8/9x^{-5/3} = 4/9x^{-5/3}(x-2)#

Points of inflection occur where f''(x) crosses the x-axis, which happens where f''(x) = 0, and there isn't a double root.

In the case of this problem, this occurs at #x=2#.

By substituting numbers around 2, we can easily see that

#f''(x) > 0, x >2# and #f''(x) < 0, x <2#

Therefore, your answer is:

There is a point of inflection at #x=2#. The graph is concave down for #x < 2#, and concave up for #x > 2#