# How do you determine if a function is concave up/ concave down if tanx+2x on (-pi/2, pi/2)?

Jul 4, 2015

Investigate the sign of the second derivative.

#### Explanation:

Let: $y = \tan x + 2 x$ on $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$

$y ' = {\sec}^{2} x + 2$ (For finding $y ' '$, remember that ${\sec}^{2} x = {\left(\sec x\right)}^{2}$ so we'll use the chain rule)

$y ' ' = 2 \sec x \sec x \tan x$ (The derivative of $2$ is $0$.)

So $y ' ' = 2 {\sec}^{2} x \tan x$.

$y ' '$ is never undefined on $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$, so its only chance to change sign is at its zero(s)..

$\left\mid \sec x \right\mid \ge 1$, so ${\sec}^{2} x > 1$ So the only zero of $y ' '$ is where $\tan x = 0$, which is at $x = 0$

For $x$ in $\left(- \frac{\pi}{2} , 0\right)$, $\tan x$ is negative, so $y ' '$ is negative and the graph of the function is concave down.

For $x$ in $\left(0 , \frac{\pi}{2}\right)$, $\tan x$ is positive, so $y ' '$ is positive and the graph of the function is concave up.

The function is concave down on $\left(- \frac{\pi}{2} , 0\right)$, and it is

concave up on $\left(0 , \frac{\pi}{2}\right)$.

The point $\left(0 , 0\right)$ is an inflection point.