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Math Help - Graphing Exponents and logs.

  1. #1
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    Graphing Exponents and logs.

    Hi, I had two question assigned to me, both of which stumped me if some one would help it would be great.

    Sketch a graph of the following and then fill in the table below:

    f(x)=3x
    g(x)=(3^2x)-3
    h(x)=g-1(x)

    Now i cant sketch them but for f and g i sketch them by imputing values for x and graph but for H I don't know how to graph that? it seems like a inverse function of G(x) but i cant quiet wrap my head around it.

    The table asks for each function state

    1) domain - All real numbers
    2)range - All real numbers
    3)asymptotes- i wrote no asymptotes for any
    4)y intercept - 1)x=0 2)x=0.5 3) didn't get this far
    5)x intercept - didn't get this far

    Thanks in advance!
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  2. #2
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    Re: Graphing Exponents and logs.

    You can rewrite h(x) so that it doesn't contain g. h(x)= (3^{2x})-3-x
    When sketching h(x) it is similar to your graph of g(x) but just subtracting x, redraw your sketch of g(x) but subtracting larger values of x as you move long the real line.

    The y intercept occurs at x=0, your intercept for g(x) is incorrect.
    The x intercept occurs at y=0, the x intercept might not exist, you can tell from your sketches.
    Thanks from Gurp925
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  3. #3
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    Re: Graphing Exponents and logs.

    My mistake, H(x) = (g^-1)(x) not g-1(x)
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  4. #4
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    Re: Graphing Exponents and logs.

    Anyone? how do i do this is its ^ the above comment
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  5. #5
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    Re: Graphing Exponents and logs.

    g(x)=y=3^{2x}-3

    y+3=3^{2x}

    ln(y+3)=ln3^{2x}

    ln(y+3)=2xln3

    \frac{ln(y+3)}{2ln3}=x

    So then g^{-1}(x)=\frac{ln(x+3)}{2ln3}
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