Finding Critical Points in g(x) = 4x - tanx

[QuarkCharmer]

Homework Statement

g(x) = 4x - tanx

Homework Equations

The Attempt at a Solution

$$g(\theta) = 4\theta - tan\theta$$
$$g'(\theta)=4-sec^{2}\theta$$
$$g'(\theta)=\frac{4cos^{2}\theta-1}{cos^{2}\theta}$$
$$g'(\theta)=\frac{-4sin^{2}\theta}{cos^{2}\theta}$$
$$-4tan^{2}\theta=0$$
$$-4(tan\theta)(tan\theta)=0$$
$$\theta = arctan0$$
$$\theta = k\pi$$

So, that lists the zeros of the derivative if I am correct. I believe the other critical points are pi/2 + kpi, where tan is undefined. That is all of the critical points correct?

 

[grey_earl]

You made an error:

...
$$g'(\theta)=\frac{4cos^{2}\theta-1}{cos^{2}\theta}$$
$$g'(\theta)=\frac{-4sin^{2}\theta}{cos^{2}\theta}$$
...

We have cos² θ + sin² θ = 1, but 4 cos² θ - 1 = 4 (1 - sin² θ) - 1 = 3 - 4 \sin² θ.

Instead you should simply set

4 cos² θ - 1 = 0,

which gives you cos θ = ±½.

 

[QuarkCharmer]

Thank you, yeah I caught that right after I posted. Now I am stuck on another problem. I suppose this should more or less be posted in the pre-calc/trig section.

Finding the critical points of:
$$f(x)=2cos(x)+sin(2x)$$
$$f(x)=2cos(x)+2sin(x)cos(x)$$

$$f'(x)=-2sin(x)+2(cos^{2}(x)-sin^{2}(x))$$
$$f'(x)=-2sin(x)+2cos^{2}(x)-2sin^{2}(x)$$

I just know there is a pythag. identity in there somewhere to aid in solving for 0.

 

[grey_earl]

Well, just express the cosine by the sine using the Pythagorean identity to get f'(x) = -2 sin x + 2 - 4 sin² x, which is a polynomial in z = sin x with roots z = -1/4 ± 3/4. Then you calculate the x corresponding to those values of sin x. That's a standard trick to solve trigonometric equations - if you have odd powers of only one trigonometric function (here sin), express the other (here cos) and you get a polynomial in the first trig function.