Domain of a function?

**Kilo** · September 13th 2008, 12:59 PM

How do you find the domain of a fuction such as f(x)=(x+sin x)/cos x?

**o_O** · September 13th 2008, 02:21 PM

Hint: For what values of $x$ will cause the denominator be equal to 0? Remember, you can't divide by 0.

**Shyam** · September 13th 2008, 02:29 PM

Originally Posted by Kilo

How do you find the domain of a fuction such as f(x)=(x+sin x)/cos x?

since denominator, cos x is zero when $x= \frac{n \pi}{2}, n\in \mathbb{I}$

So, domain is all real values except this value.

Domain = { $x\in \mathbb{R}/ x\ne\frac{n \pi}{2}, n\in \mathbb{I}$ }

or you can write this way:

Domain $= \mathbb{R} - \frac{n \pi}{2}, n\in \mathbb{I}$

**o_O** · September 13th 2008, 02:36 PM

Originally Posted by Shyam

since denominator, cos x is zero when $x= \frac{n \pi}{2}, n\in \mathbb{I}$

So, domain is all real values except this value.

Domain = { $x\in \mathbb{R}/ x\ne\frac{n \pi}{2}, n\in \mathbb{I}$ }

or you can write this way:

Domain $= \mathbb{R} - \frac{n \pi}{2}, n\in \mathbb{I}$

Not quite. For example, take n = 2. $\cos \left(\frac{2\pi}{2}\right) = \cos \pi = -1$

I think you mean: $x = \frac{\pi}{2} + n\pi \ \ n \in \mathbb{Z}$

**Kilo** · September 13th 2008, 03:17 PM

Originally Posted by Kilo

How do you find the domain of a fuction such as f(x)=(x+sin x)/cos x?

Okay, there's another problem that basically is this one....but a wee bit different....

f(x)=(x+sin x)/cos x, -3pi/2 (less than equal to) x (less than equal to) 3pi/2

Yeah, now I'm really confused....

**Moo** · September 14th 2008, 01:53 AM

Originally Posted by Kilo

Okay, there's another problem that basically is this one....but a wee bit different....

f(x)=(x+sin x)/cos x, -3pi/2 (less than equal to) x (less than equal to) 3pi/2

Yeah, now I'm really confused....

For what x between $-\frac{3 \pi}{2}$ and $\frac{3 \pi}{2}$ is $\cos(x)=0$ ?

The formula is as follows :
$\cos(a)=\cos(b) \implies \text{either } \left\{\begin{array}{ll} a=b+2k \pi \\ a=-b+2k' \pi \end{array} \right. \text{ where k and k' } \in \mathbb{Z}$

$\cos \left(\tfrac \pi 2\right)=0$ so in fact, it will give $\cos(x)=0 \Longleftrightarrow x=\frac \pi 2+k \pi$ (which is o_O's formula)

This will give you all the possibilities for $-\frac{3 \pi}{2} \le x \le \frac{3 \pi}{2}$ , that is... try to fit values for k such that x remains between these two values.

**Shyam** · September 14th 2008, 08:09 AM

Originally Posted by o_O

Not quite. For example, take n = 2. $\cos \left(\frac{2\pi}{2}\right) = \cos \pi = -1$

I think you mean: $x = \frac{\pi}{2} + n\pi \ \ n \in \mathbb{Z}$

Oh, yes, I realised it later. There was a mistake. Actually it should be,

Domain={ $x\in \mathbb{R}/ x\ne\frac{(2n-1) \pi}{2}, n\in \mathbb{I}$ }

or you can say

Domain= $\mathbb{R} - \frac{(2n-1) \pi}{2}, n\in \mathbb{I}$

because, cos x is zero for odd multiples of $\frac{\pi}{2}$

**Shyam** · September 14th 2008, 08:18 AM

Originally Posted by Kilo

Okay, there's another problem that basically is this one....but a wee bit different....

f(x)=(x+sin x)/cos x, -3pi/2 (less than equal to) x (less than equal to) 3pi/2

Yeah, now I'm really confused....

you see, cos x is 0 when x =odd multiple of $\frac{\pi}{2}$

so, $x \ne \frac{-3\pi}{2},\frac{-\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}$

Now, you try to write the domain.

**Kilo** · September 14th 2008, 01:48 PM

Thanks everyone! It's starting to make sense to me now!

**log(xy)** · September 15th 2008, 07:11 AM

i know this thread is kinda done with, but cant the domain really just be any real number? in algebra we know that the domain of a function can be any real number, so long as it has only one range, but thats besides the point of the fact that shouldnt the answer be any real number?

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