How do you determine the intervals where $f(x)=x^(2/3)+3$ is concave up or down?

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Answer 1 · 2016-10-25T15:16:21

Investigate the sign of the second derivative.

$f(x) = x^(2/3)+3$

The domain of $f$ is all real numbers.

$f'(x) = 2/3x^(-1/3)$

$f''(x) = -2/9 x^(-4/3) = -2/(9root(3)x^4)$

$f''$ is never $0$ and is undefined at $x=0$ .

So the only place where $f''$ might change signs is $x=0$

On $(-oo,0)$ , we see that $f''$ is negative, so $f$ is concave down.

On $(0,oo)$ , we see again that $f''$ is negative, so $f$ is concave down.

Because the concavity does not change, there is no inflection point.

By the way, here is the graph of $f(x)$ .

graph{y=x^(2/3)+3 [-12.62, 12.69, -2.98, 9.68]}

How do you determine the intervals where f(x)=x^(2/3)+3f(x)=x^(2/3)+3 is concave up or down?