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Thread: Inverse Functions

  1. #1
    Newbie Tom G's Avatar
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    Inverse Functions

    I've got two functions [f(x) and g(x)] which I need to find the inverse of [f-(x) and g-(x)], can someone please explain the steps taken when finding the inverse of funtions.

    1. f(x)= the square root of 'x' (i don't have a square root button on my keyboard)

    2. g(x)= 2x+1

    I would appreciate if someone could show each step they have taken to get the answer.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Tom G View Post
    I've got two functions [f(x) and g(x)] which I need to find the inverse of [f-(x) and g-(x)], can someone please explain the steps taken when finding the inverse of funtions.

    1. f(x)= the square root of 'x' (i don't have a square root button on my keyboard)

    2. g(x)= 2x+1

    I would appreciate if someone could show each step they have taken to get the answer.
    The general method is to take a function $\displaystyle y = f(x)$, reverse the roles of x and y ($\displaystyle x = f(y)$, then solve this equation for y ($\displaystyle y = g(x) = f^{-1}(x)$.)

    When doing this you need to be careful about domains and ranges. (The range of the function is the domain of the inverse function and visa versa.)

    For example:
    1) $\displaystyle y = f(x) = \sqrt{x}$
    Switch the roles of x and y: $\displaystyle x = \sqrt{y}$
    Solve for y: $\displaystyle y = x^2$

    Thus $\displaystyle f^{-1}(x) = x^2$.

    Note, though, that since f(x) is only defined on $\displaystyle [0, \infty )$ that the inverse function is only defined on $\displaystyle [0, \infty )$ rather than $\displaystyle (-\infty, \infty )$.


    2) $\displaystyle g(x) = 2x + 1$.
    You do this one. I get $\displaystyle g^{-1}(x) = \frac{x - 1}{2}$.

    -Dan
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  3. #3
    MHF Contributor red_dog's Avatar
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    If $\displaystyle f:A\to B$ and $\displaystyle f(x)=y$ then $\displaystyle f^{-1}:B\to A$ and $\displaystyle f^{-1}(y)=x$.
    So, to find $\displaystyle f^{-1}$ you have to find $\displaystyle y$ as a function of $\displaystyle x$.

    1) $\displaystyle f:[0,\infty)\to[0,\infty),f(x)=\sqrt{x}$.
    $\displaystyle \sqrt{x}=y$
    Square both sides:
    $\displaystyle x=y^2$ So, $\displaystyle f^{-1}(y)=y^2$
    Now, change the letter for the variable:
    $\displaystyle f^{-1}:[0,\infty)\to[0,\infty), \ f^{-1}(x)=x^2$

    Now, can you solve 2)?
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  4. #4
    Newbie Tom G's Avatar
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    Thanks, I've now managed to do g(x) and got the same answer as topsquark.

    I've got:

    g(x)=2x+1

    x=2y+1
    x-1=2y
    (x-1)/2=y

    (x-1)/2=g-(x)
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