# For what values of x is f(x)=-x^4+4x^3-2x^2-x+5 concave or convex?

Jun 1, 2016

$x < \frac{1}{3} \left(3 - \sqrt{6}\right) \to \text{convexity}$
$\frac{1}{3} \left(3 - \sqrt{6}\right) < x < \frac{1}{3} \left(3 + \sqrt{6}\right) \to \text{concavity}$
$\frac{1}{3} \left(3 + \sqrt{6}\right) < x \to \text{convexity}$

#### Explanation:

For $f \left(x\right)$ the local concavity/convexity is given by the sign of

$\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} \left(x\right)$.

For $f \left(x\right) = - {x}^{4} + 4 {x}^{3} - 2 {x}^{2} - x + 5$
we have $\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} \left(x\right) = - 4 + 24 x - 12 {x}^{2}$.

If $\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} \left(x\right) < 0$ the local curvature is qualified as convex
otherwise if $\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} \left(x\right) > 0$ concave.

Extracting the roots of

$- 4 + 24 x - 12 {x}^{2} = 0$

we get

$\left\{\begin{matrix}x = \frac{1}{3} \left(3 - \sqrt{6}\right) \\ x = \frac{1}{3} \left(3 + \sqrt{6}\right)\end{matrix}\right.$

$x < \frac{1}{3} \left(3 - \sqrt{6}\right) \to \text{convexity}$
$\frac{1}{3} \left(3 - \sqrt{6}\right) < x < \frac{1}{3} \left(3 + \sqrt{6}\right) \to \text{concavity}$
$\frac{1}{3} \left(3 + \sqrt{6}\right) < x \to \text{convexity}$

blue $f \left(x\right)$, pink $\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} \left(x\right)$