# For what values of x is f(x)=(x-2)(x-7)(x-3) concave or convex?

Jan 22, 2016

Concave on $\left(- \infty , 4\right)$; convex on $\left(4 , + \infty\right)$

#### Explanation:

The concavity and convexity of a function are determined by the sign (positive/negative) of the second derivative.

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

In order to find the second derivative, we should first simplify the undifferentiated function by distributing.

$f \left(x\right) = \left({x}^{2} - 9 x + 14\right) \left(x - 3\right) = {x}^{3} - 12 {x}^{2} + 41 x - 42$

Now, find the first and second derivatives through a simple application of the power rule.

$f ' \left(x\right) = 3 {x}^{2} - 24 x + 41$
$f ' ' \left(x\right) = 6 x - 24$

Now, we must find the times when $6 x - 24$ is positive and when it is negative. The times when the function could shift from positive to negative or vice versa are when $6 x - 24 = 0$.

$6 x - 24 = 0$
$6 x = 24$
$x = 4$

The sign of the second derivative, and by extension, the concavity/convexity of the function, could shift only at $x = 4$. Thus, we should test points on either side of $x = 4$ to determine which concavity/convexity is present on either side.

When $m a t h b f \left(x < 4\right)$:

Test point at $x = 0$:

$f ' ' \left(0\right) = 6 \left(0\right) - 24 = - 24$

Since this is $< 0$, the entire interval $\left(- \infty , 4\right)$ will be concave.

When $m a t h b f \left(x > 4\right)$:

Test point at $x = 5$:

$f ' ' \left(5\right) = 6 \left(5\right) - 24 = 6$

Since this is $> 0$, the entire interval $\left(4 , + \infty\right)$ will be convex.

We can check the graph of $f \left(x\right)$: convexity is characterized by a $\cup$ shape, and concavity is characterized by a $\cap$ shape. The concavity of the graph should shift at $x = 4$.

graph{x^3-12x^2+41x-42 [-2, 10, -30, 15]}