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Thread: domain of a composite function

  1. #1
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    domain of a composite function

    Say i am given 2 functions , f(x) and g(x) with their respective domains , is it possible for me to know the domain of f o g without evaluating function f o g ?

    Example :

    $\displaystyle f(x)=\sqrt{x-1}$ , $\displaystyle x\geq 1$

    $\displaystyle g(x)=\frac{1}{x-1}$ , $\displaystyle x\in R$ , $\displaystyle x\neq 1$


    $\displaystyle fg(x)=\sqrt{\frac{1}{x-1}-1}$

    i see that the expression under the square roots must be >=0 , so

    $\displaystyle 1\leq x \leq 2$ but since $\displaystyle x\neq 1$ so the domain is $\displaystyle 1<x\leq 2$

    Thats how i would do it . Is there an easier way ?
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  2. #2
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    Quote Originally Posted by hooke View Post
    Say i am given 2 functions , f(x) and g(x) with their respective domains , is it possible for me to know the domain of f o g without evaluating function f o g ?

    Example :

    $\displaystyle f(x)=\sqrt{x-1}$ , $\displaystyle x\geq 1$

    $\displaystyle g(x)=\frac{1}{x-1}$ , $\displaystyle x\in R$ , $\displaystyle x\neq 1$


    $\displaystyle fg(x)=\sqrt{\frac{1}{x-1}-1}$

    i see that the expression under the square roots must be >=0 , so

    $\displaystyle 1\leq x \leq 2$ but since $\displaystyle x\neq 1$ so the domain is $\displaystyle 1<x\leq 2$

    Thats how i would do it . Is there an easier way ?
    x will be in the domain of fog if it is in the domain of g and g(x) is in the domain of f.

    Here, $\displaystyle g(x)= \frac{1}{x- 1}$ so its domain is all real numbers except 1. $\displaystyle f(x)= \sqrt{x- 1}$ which has domain $\displaystyle x\ge 1$.

    The domain of fog will be all x, except 1, such that $\displaystyle g(x)= \frac{1}{x- 1}\ge 1$. If x> 1, x-1> 0 so, multiplying on both sides by the positive number x- 1, $\displaystyle 1\ge x- 1$ or $\displaystyle x\le 2$. If x< 1, x- 1< 0 so, multiplying both sides by the negative number x- 1, $\displaystyle 1\le x- 1$ or $\displaystyle x\ge 2$. Of course, we can't have "x< 1" and "$\displaystyle x\ge 2$" so that last case is impossible. We must have x> 1 and $\displaystyle x\le 2$ as you say.
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