Critical points of a trig function

[colonelAP]

Homework Statement

g(x)=x - sin (pi*x) at [0,2]

Homework Equations

all values where g'(x)= 0 (or non-existant) are critical points

The Attempt at a Solution

g'(x)=1 - ╥(cos(╥X))
1/╥=cos(╥X)
cos^(-1) (1/╥) = ╥X
X=(cos^(-1) (1/╥))/╥
x=.397 and x=? 2nd point

 

[GuiltySparks]

Remember your unit circle. Taking the inverse cosine of (1/pi) points to one of two possible triangles that lie on this circle. From the argument, we know that the x-component of this circle will be 1; the hypotenuse will be pi. Remember that the hypotenuse will always be positive but the x- or y-components can be either negative or positive. So, since the x-component is positive, we can have either a positive or a negative y-value to realize our two triangles. So, the angle is +/- 1.25 radians.

x = +/- 0.397

 

[SammyS]

Homework Statement

g(x)=x - sin (pi*x) at [0,2]

Homework Equations

all values where g'(x)= 0 (or non-existant) are critical points

The Attempt at a Solution

g'(x)=1 - ╥(cos(╥X))
1/╥=cos(╥X)
cos^(-1) (1/╥) = ╥X
X=(cos^(-1) (1/╥))/╥
x=.397 and x=? 2nd point

Hello colonelAP. Welcome to PF !

To help us respond to questions, you should include the complete problem in the text of your post, even if you wrote a portion of what you're looking for in the title of your thread.

I gather that you are to find all the critical points in the interval [0. 2] for the function

f(x)=x-\sin(\pi x)\,.​

What is the period of \cos(\pi x)\,?

Is \cos(\pi x) an even function, or is it an odd function?