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Thread: Inverse function

  1. #1
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    Inverse function

    Hi, I could really use some help with this. I couldn't understand the book's way of solving the problem.

    Find the inverse function of the following, also state the domain and range of the original functions:

    a) g(x)=3+e^x
    b) h(x)=log(x-1)

    Thanks in advance.
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    Re: Inverse function

    Quote Originally Posted by FatimaA View Post
    Hi, I could really use some help with this. I couldn't understand the book's way of solving the problem.

    Find the inverse function of the following, also state the domain and range of the original functions:

    a) g(x)=3+e^x
    b) h(x)=log(x-1)

    Thanks in advance.
    Are you familiar with the general procedure for finding inverse functions? Interchange the $\displaystyle x$ and the $\displaystyle y$ and solve for $\displaystyle y$. This may involve taking the logarithm of both sides, or exponentiating both sides.

    So for (a), we start with $\displaystyle y = 3+e^x$ and we switch the $\displaystyle x$'s and $\displaystyle y$'s, producing

    $\displaystyle x = 3+e^y$.

    Now, can you solve the above for $\displaystyle y$?
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    Re: Inverse function

    Yeah I'm familiar with interchanging the x and y but I'm pretty sure my teacher wants us to take the logarithm of both sides, which I am really, really confused about. I don't know what you mean by exponentiating.

    For y= 3+e^x I thought you were supposed to get the x on one side and then switch the variables?
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    Re: Inverse function

    Quote Originally Posted by FatimaA View Post
    For y= 3+e^x I thought you were supposed to get the x on one side and then switch the variables?
    Yes, you could solve for $\displaystyle x$ and then switch the variables after; that accomplishes the same thing. So the problem is how to solve the equation $\displaystyle y = 3+e^x$ for $\displaystyle x$. Begin by isolating the exponential term:

    $\displaystyle y=3+e^x\quad\Rightarrow\quad e^x=y-3$

    Now take the natural logarithm of both sides:

    $\displaystyle e^x=y-3\quad\Rightarrow\quad\ln e^x=\ln(y-3)$

    And $\displaystyle \ln e^x$ is just $\displaystyle x$ (i.e., these are inverse operations). So we are left with

    $\displaystyle x = \ln(y-3)$.


    For the second one, I assume that the base is 10. You can cancel out the operation of the logarithm by raising 10 to the power of each side:

    $\displaystyle y=\log(x-1)\quad\Rightarrow\quad10^y=10^{\log(x-1)}\quad\Rightarrow\quad10^y=x-1$.

    Basically, you are converting the equation from logarithmic form $\displaystyle \log_bx=a$ to the exponential form $\displaystyle b^a=x$.
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  5. #5
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    Re: Inverse function

    Thanks! Actually I found the procedure in my notes lol it was there all along but now I can check my work. I did exactly what you did, but this way:

    1) y=3+e^x

    2) y-3=e^x

    3) ln(y-3)=ln(e^x)

    4) x=ln(y-3)

    5) g^-1(x)=ln(x-3)

    I got the same answer as you for the second one as well. I did it pretty much the same way:

    1) y=log(x-1)

    2) 10^y=10^log(x-1)

    3) 10^y=x-1

    4) (10^y)+1=x

    5) h^-1(x)=(10^x)+1

    Obviously I took many more steps but it looks like we both got the same thing.
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