# For what values of x is f(x)= 2x^3-9x  concave or convex?

Jan 6, 2017

$x = \pm \sqrt{\frac{3}{2}}$. Read on to see which is which.

I try not to confuse myself over "concave" vs. "convex". Instead I think about it as concave up or concave down.

The first derivative equals $0$ at an extremum, which may be either an inflection point, concave up, or concave down.

It is the second derivative at each of these points that tells you which of these three they are. Positive, if concave up, and negative, if concave down.

$f ' \left(x\right) = 6 {x}^{2} - 9$

(power rule; $\frac{d}{\mathrm{dx}} \left[{x}^{n} = n {x}^{n - 1}\right]$)

$f ' ' \left(x\right) = 12 x$

For us to find where the extrema are:

$0 = 6 {x}^{2} - 9$

$\implies {x}^{2} = \frac{9}{6} = \frac{3}{2}$

$\implies x = \pm \sqrt{\frac{3}{2}}$

And to find which one is concave up/down or an inflection point, we take $f ' ' \left(\pm \sqrt{\frac{3}{2}}\right)$.

$f ' ' \left(\sqrt{\frac{3}{2}}\right) = 12 \left(\sqrt{\frac{3}{2}}\right) > 0$

$f ' ' \left(- \sqrt{\frac{3}{2}}\right) = 12 \left(- \sqrt{\frac{3}{2}}\right) < 0$

Thus, $f \left(x\right)$ is concave up (a minimum) at $x = \sqrt{\frac{3}{2}}$ and concave down (a maximum) at $x = - \sqrt{\frac{3}{2}}$.

Indeed, $f \left(x\right) = 2 {x}^{3} - 9 x$ has features of ${x}^{3}$ and $- x$. We can expect it to look like a cubic function that has a $y = - x$ close to the origin, like so:

graph{2x^3 - 9x [-10,10, -10.14, 10.13]}