Results 1 to 3 of 3
Like Tree2Thanks
  • 1 Post By skeeter
  • 1 Post By jakncoke

Thread: Domain and Range

  1. #1
    Junior Member
    Joined
    Oct 2012
    From
    Australia
    Posts
    43

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702

    Re: Domain and Range

    (a) domain ...

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

    $\displaystyle e^x \ge 1$

    $\displaystyle x \ge 0$

    range ...

    $\displaystyle \cos(whatever)$ is between what two values?


    (b) domain ...

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

    range ...

    $\displaystyle 0 < \frac{1}{x^2+1} < \, ?$
    Thanks from NettieL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member jakncoke's Avatar
    Joined
    May 2010
    Posts
    387
    Thanks
    80

    Re: Domain and Range

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

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

    So for ur first example

    1) $\displaystyle cos(\sqrt{e^x - 1}) $. cos(x) is defined for all real numbers, but $\displaystyle \sqrt{e^x - 1} $ is real only for $\displaystyle e^x - 1 \geq 0 $ (because $\displaystyle \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 $\displaystyle f(x) = \frac{1}{x-1} $. Since we said the domain of f was all real numbers except 1. We see that pluggin in $\displaystyle x > 1 $, for $\displaystyle 1 < x < 2 $ makes the denominator in $\displaystyle f(x) = \frac{1}{x-1} $ makes the denominator < 1, which means the function it it self can become as large as $\displaystyle \infty $ So for $\displaystyle 1 < x < 2 $ our range is $\displaystyle (1, \infty) $. Now take $\displaystyle x \geq 2 $ So our range can take on values from $\displaystyle (0, 1] $ So our range for x > 1 is $\displaystyle (0, 1] $ plus $\displaystyle (1, \infty) $ which means range is $\displaystyle (0, \infty ) $Now since we did x > 1, now we try x < 1, so for x < 1, we see that for if $\displaystyle 0 \geq x \< 1 $ our denominator takes on the values in the interval $\displaystyle [-\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 $\displaystyle (0, \infty) plus (-\infty, -1] plus [-1, 0)$, which means that our whole range is $\displaystyle (-\infty, 0) \cup (0, \infty) $, basically ur function can take on every value except 0.
    Last edited by jakncoke; Nov 18th 2012 at 03:38 PM.
    Thanks from NettieL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: Sep 16th 2009, 06:25 AM
  2. Domain and Range of f(x,y)
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Oct 1st 2008, 09:20 PM
  3. Range and Domain
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: Sep 28th 2008, 04:42 PM
  4. Domain and range
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Feb 27th 2008, 03:11 AM
  5. [SOLVED] domain & range
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Sep 3rd 2007, 06:04 AM

Search Tags


/mathhelpforum @mathhelpforum