# For what values of x is f(x)= 4x^3-12x^2  concave or convex?

Jan 20, 2016

Concave on $\left(- \infty , 1\right)$; convex on $\left(1 , + \infty\right)$

#### Explanation:

The convexity and concavity of a function and determined by the sign of the second derivative.

• If $f ' ' \left(a\right) > 0$, then $f \left(x\right)$ is convex at $x = a$.
• If $f ' ' \left(a\right) < 0$, then $f \left(x\right)$ is concave at $x = a$.

First, find the second derivative.

$f \left(x\right) = 4 {x}^{3} - 12 {x}^{2}$
$f ' \left(x\right) = 12 {x}^{2} - 24 x$
$f ' ' \left(x\right) = 24 x - 24$

The second derivative could change signs whenever it is equal to $0$. Find that point by setting the second derivative equal to $0$.

$24 x - 24 = 0$
$x = 1$

The convexity/concavity could shift only at this point. Thus, from here, we can determine on which intervals the function will be uninterruptedly convex or concave.

Use test points around $x = 1$:

When $m a t h b f \left(x < 1\right)$:

$f ' ' \left(0\right) = - 24$

Since this is $< 0$, the function is concave on the interval $\left(- \infty , 1\right)$.

When $m a t h b f \left(x > 1\right)$:

$f ' ' \left(2\right) = 24$

Since this is $> 0$, the function is convex on the interval $\left(1 , + \infty\right)$.

Always consult a graph of the original function when possible:

graph{4x^3-12x^2 [-2 5, -19.9, 5.77]}

The concavity does seem to shift around the point $x = 1$. When $x < 1$, the graph points downward, in the $\cap$ shape characteristic of concavity. When $x > 1$, the graph points upward in the $\cup$ shape characteristic of convexity.