How do i find the range of this function?

[jd12345]

Homework Statement

Find the range of y = (cos²x - 1) / ( cos²x + cos x )

Homework Equations

The Attempt at a Solution

Well i tried the usual method. I cross multiplied and got a quadratic equation in cos x. Then it should have discriminant greater than zero so in the end i get (y-2)² > 0 which is true for all y. So range is R? ( I don't have the answer)

 

[HallsofIvy]

Is the formula given correct? Do you really mean
[tex]y= \frac{cos(x)-1}{2cos(x)}[/tex]?
That is equal to
[tex]\frac{1}{2}- \frac{1}{2cos(x)}[/tex]
Since 2cos(x) is never larger than 2 or less than -2, 1/(2cos(x)) is always larger than 1/2 or less than -1/2, never between.

 

[Mentallic]

Try factoring the equation, you should get a nice simplification.

But when you simplify you need to keep in mind that if you have an expression of the form [tex]y=\frac{ab}{b}[/tex] then when you simplify to y=a, you need to keep in mind that [itex]b\neq 0[/itex], so you would have to exclude this value if it's in the range.

 

[Infinitum]

The problem with your method is that since cos is a restricted function, its own range is only between [-1,1]. To use your method, the equation should be in tan or cot functions since they have range R. Or you can restrict the answer you get accordingly to the range of cos.

An easier method would be to factorize the equation since it results in a very simple equation.

Edit : just noticed Mentallic already posted the easier way.

 

[SammyS]

Homework Statement

Find the range of y = (cos²x - 1) / ( cos²x + cos x )

Homework Equations

The Attempt at a Solution

Well i tried the usual method. I cross multiplied and got a quadratic equation in cos x. Then it should have discriminant greater than zero so in the end i get (y-2)² > 0 which is true for all y. So range is R? ( I don't have the answer)

Simplify [itex]\displaystyle \frac{\cos^2(x)-1}{\cos^2(x)+\cos(x)}\ .[/itex]

Factor the numerator & the denominator, then cancel, keeping in mind what Mentallic & Infinitum mentioned.