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

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Answer 1 · 2018-01-08T18:30:15

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

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

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 $x > 2$

How do you find points of inflection and determine the intervals of concavity given y=x^(1/3)(x-4)y=x13(x−4)y=x^(1/3)(x-4)?