# Is f(x)=tanx concave or convex at x=pi/3?

Jun 11, 2016

Convex

#### Explanation:

In order to find the concavity of a function at some point $x$, you need to take the second derivative of the function. Then, you need to find the sign of the second derivative:

• If $f ' ' \left(x\right) > 0$, the function is convex.
• If $f ' ' \left(x\right) < 0$, the function is concave.

For $f \left(x\right) = \tan x$:
$f ' \left(x\right) = {\sec}^{2} x$
$f ' ' \left(x\right) = 2 \sec x \tan x \left(\sec x\right) = 2 {\sec}^{2} x \tan x$

Now we have to evaluate this at $x = \frac{\pi}{3}$:
$f ' ' \left(\frac{\pi}{3}\right) = 2 {\sec}^{2} \left(\frac{\pi}{3}\right) \tan \left(\frac{\pi}{3}\right)$

Before we go about this, remember that we're only concerned about the sign of the second derivative, not the actual value. We don't actually need to compute $2 {\sec}^{2} \left(\frac{\pi}{3}\right) \tan \left(\frac{\pi}{3}\right)$ - we just have to find out if it's negative or positive.

Take a close look at this expression. $\frac{\pi}{3}$ is an angle in the first quadrant (it's in between $0$ and $\frac{\pi}{2}$), and all the trig functions are positive in the first quadrant. That means all three of the pieces we have here are positive:

$\underbrace{2} \underbrace{{\sec}^{2} \left(\frac{\pi}{3}\right)} \underbrace{\tan} \left(\frac{\pi}{3}\right)$
$+$$\textcolor{w h i t e}{X X} +$$\textcolor{w h i t e}{X X X} +$

And when you multiply positive numbers, you get a positive result. Therefore, we can conclude that $f ' ' \left(\frac{\pi}{3}\right) > 0$, so the function is convex at that point. You can confirm this by looking at the function's graph.