# For what values of x is f(x)= -x^4-9x^3+2x+4  concave or convex?

Mar 15, 2018

Convex for $x \in \left(- \frac{9}{2} , 0\right)$

Concave for $x \in \left(- \infty , - \frac{9}{2}\right) \cup \left(0 , \infty\right)$

#### Explanation:

A function is convex where its second derivative $f ' ' > 0$, and concave where its second derivative $f ' ' < 0$

The second derivative is the derivative of the first derivative .i.e.

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

$f ' \left(x\right) = - 4 {x}^{3} - 27 {x}^{2} + 2$

f''(x)=f'(f'(x)=f'(-4x^3-27x^2+2)=-12x^2-54x

Convex:

$- 12 {x}^{2} - 54 x > 0$

$x \left(- 12 x - 54\right) > 0$

$x > 0$

$- 12 x - 54 > 0$

$- 12 x > 54$

$x < - \frac{54}{12}$

$x < - \frac{9}{2}$

$x < 0$

$- 12 x - 54 < 0$

$x > \frac{54}{-} 12$ , $x > - \frac{9}{2}$

Convex for $x \in \left(- \frac{9}{2} , 0\right)$

Concave:

$- 12 {x}^{2} - 54 x < 0$

$x \left(- 12 x - 54\right) < 0$

$x < 0$

$- 12 x - 54 < 0$ , $x > - \frac{9}{2}$

$x > 0$

$- 12 x - 54 > 0$ , $x < - \frac{9}{2}$

Concave for $x \in \left(- \infty , - \frac{9}{2}\right) \cup \left(0 , \infty\right)$