# For what values of x is f(x)=3x^3-7x^2-5x+9 concave or convex?

Jun 5, 2016

$f$ is concave (concave down) on $\left(- \infty , \frac{7}{9}\right)$ and is convex (concave up) on $\left(\frac{7}{9} , \infty\right)$.

#### Explanation:

The convexity and concavity of the function $f$ can be determined by looking at the sign of the second derivative:

• If $f ' ' > 0$, then $f$ is convex.
• If $f ' ' < 0$, then $f$ is concave.

To find the function's second derivative, use the power rule.

$f \left(x\right) = 3 {x}^{3} - 7 {x}^{2} - 5 x + 9$

$f ' \left(x\right) = 9 {x}^{2} - 14 x - 5$

$f ' ' \left(x\right) = 18 x - 14$

So, the convexity and concavity are determined by the sign of $f ' ' \left(x\right) = 18 x - 14$.

The second derivative equals $0$ when $18 x - 14 = 0$, which is at $x = \frac{7}{9}$.

When $x > \frac{7}{9}$, $f ' ' \left(x\right) > 0$, so $f \left(x\right)$ is convex on $\left(\frac{7}{9} , \infty\right)$.

When $x < \frac{7}{9}$, $f ' ' \left(x\right) < 0$, so $f \left(x\right)$ is concave on $\left(- \infty , \frac{7}{9}\right)$.