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Math Help - Domain and Range

  1. #1
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    Domain and Range

    Hi, just wonderin if I could have some help with these two questions: each of them asks to find the domain and the range using calculus techniques, and as I am reveiwing this after doing this 12 weeks ago, I am a bit hazy as to what's required.
    (a)f(x)=cos sqrt(e^x-1)

    (b) g(x)=1//(x^2+1)

    if anyone could help and explain fully, that would be greatly appreciated
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  2. #2
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    Re: Domain and Range

    (a) domain ...

    e^x - 1 \ge 0 (why?)

    e^x \ge 1

    x \ge 0

    range ...

    \cos(whatever) is between what two values?


    (b) domain ...

    x^2+1 > 0 for all x \in \mathbb{R} , so what is the domain?

    range ...

    0 < \frac{1}{x^2+1} < \, ?
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  3. #3
    Senior Member jakncoke's Avatar
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    Re: Domain and Range

    Basically finding the domain means that all the places in  \mathbb{R} where the function is defined. So for example

    1) Find the domain of  f(x) = \frac{1}{x-1} . The domain is All real numbers except 1. Because  f(1) = \frac{1}{1-1} and u cannot divide by 0.

    So for ur first example

    1)  cos(\sqrt{e^x - 1}) . cos(x) is defined for all real numbers, but  \sqrt{e^x - 1} is real only for  e^x - 1 \geq 0 (because  \sqrt(negative num) is not a real number, its a complex number).


    As for the range, range is basically all the values ur function can take if u plug in x's from ur domain only.

    For example
    2) Find the Range of  f(x) = \frac{1}{x-1} . Since we said the domain of f was all real numbers except 1. We see that pluggin in  x > 1 , for  1 < x  < 2 makes the denominator in  f(x) = \frac{1}{x-1} makes the denominator < 1, which means the function it it self can become as large as  \infty So for  1 < x < 2 our range is (1, \infty) . Now take x \geq 2 So our range can take on values from  (0, 1] So our range for x > 1 is  (0, 1] plus  (1, \infty) which means range is  (0, \infty ) Now since we did x > 1, now we try x < 1, so for x < 1, we see that for if  0 \geq x \< 1 our denominator takes on the values in the interval  [-\infty, -1 ] . and if x < 0, we get a small negative denominator, and the function takes on every value from [-1, 0). So our range is (0, \infty) plus (-\infty, -1] plus [-1, 0), which means that our whole range is (-\infty, 0) \cup (0, \infty) , basically ur function can take on every value except 0.
    Last edited by jakncoke; November 18th 2012 at 03:38 PM.
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