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Math Help - Domain of a function?

  1. #1
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    Domain of a function?

    How do you find the domain of a fuction such as f(x)=(x+sin x)/cos x?
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  2. #2
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    Hint: For what values of x will cause the denominator be equal to 0? Remember, you can't divide by 0.
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    Quote Originally Posted by Kilo View Post
    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}
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    Quote Originally Posted by Shyam View Post
    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}
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    Quote Originally Posted by Kilo View Post
    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....
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  6. #6
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    Quote Originally Posted by Kilo View Post
    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.
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  7. #7
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    Quote Originally Posted by o_O View Post
    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}
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  8. #8
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    Quote Originally Posted by Kilo View Post
    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.
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  9. #9
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    Thanks everyone! It's starting to make sense to me now!
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  10. #10
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    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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