# How do you find the critical numbers of g(θ) = 4 θ - tan(θ)?

Mar 23, 2017

$\theta = 0$ is an inflection point.

#### Explanation:

Extrema
$g ' \left(\theta\right) = 4 - \setminus \frac{1}{\setminus {\cos}^{2} \left(\theta\right)} = 0$
or
$\setminus {\cos}^{2} \left(\theta\right) = \setminus \frac{1}{4}$
$\setminus \cos \left(\theta\right) = \pm \setminus \frac{1}{2}$

That is
$\theta = \setminus \frac{\pi}{3}$
$\theta = \setminus \frac{5 \setminus \pi}{3}$
for $\cos \left(\theta\right) = \setminus \frac{1}{2}$
and
$\theta = \setminus \frac{2 \setminus \pi}{3}$
$\theta = \setminus \frac{4 \setminus \pi}{3}$
for $\cos \left(\theta\right) = - \setminus \frac{1}{2}$

$g ' ' \left(\theta\right) = - \left(- 2\right) {\cos}^{- 3} \left(\theta\right) \left(- \sin \left(\theta\right)\right) = - 2 {\cos}^{- 3} \left(\theta\right) \sin \left(\theta\right)$

graph{4x-tan(x) [-7, 7, -7, 7]}
graph{sin(x)/cos(x)^3 [-7, 7, -30, 30]}

As you can see on the bottom plot, and as expected, the second derivative of g changes sign at x = 0. Therefore x=0 is an inflection point. The extrema are true extrema. The second derivative does not change sign at the extrema. If it did, these would not be extrema but saddle points.