Question

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

Answer 1

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''(x)>0`, the function is convex.
• If `f''(x)<0`, the function is concave.

For `f(x)=tanx`:
`f'(x)=sec^2x`
`f''(x)=2secxtanx(secx)=2sec^2xtanx`

Now we have to evaluate this at `x=pi/3`:
`f''(pi/3)=2sec^2(pi/3)tan(pi/3)`

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 `2sec^2(pi/3)tan(pi/3)` - we just have to find out if it's negative or positive.

Take a close look at this expression. `pi/3` is an angle in the first quadrant (it's in between `0` and `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(pi/3))underbracetan(pi/3)`
`+` `color(white)(XX)+` `color(white)(XXX)+`

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