Is f(x) = (x+3)^(2/3) - 6x^3 concave or convex at x=-1?

Mar 12, 2017

$f \left(x\right)$ is decreasing and is concave up at $x = - 1$

Explanation:

The answer to this question lies in its first and second derivative.

First, lets find the first and second derivative.

If, $f \left(x\right) = {\left(x + 3\right)}^{\frac{3}{2}} - 6 {x}^{3}$

using the power rule, the first derivative is:

$f ' \left(x\right) = \left(\frac{3}{2}\right) {\left(x + 3\right)}^{\frac{1}{2}} - 6 \cdot 3 {x}^{2}$

$= \left(\frac{3}{2}\right) {\left(x + 3\right)}^{\frac{1}{2}} - 18 {x}^{2}$

Using the power rule once more, the second derivative is:

$f ' ' \left(x\right) = \left(\frac{3}{2}\right) \left(\frac{1}{2}\right) {\left(x + 3\right)}^{- \frac{1}{2}} - 18 \cdot 2 x$

$= \frac{3}{4 {\left(x + 3\right)}^{\frac{1}{2}}} - 36 x$

Now, recall that if, at an $x$-value, $a$,

$f ' \left(x\right) > 0$, $f \left(x\right)$ is increasing at $x = a$
$f ' \left(x\right) < 0$, $f \left(x\right)$ is decreasing at $x = a$

Also, recall that if, at an $x$-value, $a$,

$f ' ' \left(x\right) > 0$, $f \left(x\right)$ is concave up at $x = a$
$f ' ' \left(x\right) < 0$, $f \left(x\right)$ is concave down at $x = a$

So,
$f ' \left(- 1\right) = \left(\frac{3}{2}\right) {\left(- 1 + 3\right)}^{\frac{1}{2}} - 18 {\left(- 1\right)}^{2}$

$= \frac{3 \sqrt{2}}{2} - 18 = \frac{3 \sqrt{2} - 36}{2}$ which is clearly negative.

And, $f ' ' \left(- 1\right) = \frac{3}{4 {\left(- 1 + 3\right)}^{\frac{1}{2}}} - 36 \left(- 1\right)$

$= \frac{3}{4 \sqrt{2}} + 36$ which is clearly positive.

Thus, $f \left(x\right)$ is decreasing and is concave up at $x = - 1$