How do you determine if a function is concave up/ concave down if $tan x + 2 x$ $tanx+2x$ on $(- \frac{π}{2}, \frac{π}{2})$ $(-pi/2, pi/2)$ ?

Question

9361 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-04T03:24:24

Investigate the sign of the second derivative.

Let: $y = tan x + 2 x$ $y = tanx + 2x$ on $(- \frac{π}{2}, \frac{π}{2})$ $(-pi/2, pi/2)$

$y' = sec^2x +2$ (For finding $y''$ , remember that $sec^2x = (secx)^2$ so we'll use the chain rule)

$y'' = 2secx secx tanx$ (The derivative of $2$ is $0$ .)

So $y'' = 2sec^2x tanx$ .

$y''$ is never undefined on $(-pi/2, pi/2)$ , so its only chance to change sign is at its zero(s)..

$abs(sec x) >= 1$ , so $sec^2x > 1$ So the only zero of $y''$ is where $tanx = 0$ , which is at $x = 0$

For $x$ in $(-pi/2, 0)$ , $tanx$ is negative, so $y''$ is negative and the graph of the function is concave down.

For $x$ in $(0, pi/2)$ , $tanx$ is positive, so $y''$ is positive and the graph of the function is concave up.

The function is concave down on $(-pi/2, 0)$ , and it is

concave up on $(0, pi/2)$ .

The point $(0,0)$ is an inflection point.

How do you determine if a function is concave up/ concave down if tanx+2x tanx+2xtanx+2x on (-pi/2, pi/2)(−π2,π2)(-pi/2, pi/2)?