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Thread: f*g(x) and f/g(x) homework check

  1. #1
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    f*g(x) and f/g(x) homework check

    hey guys,
    I just want to check if i did these questions right, ym textbook doesn't have any examples of them, so there's no way for me to check

    a) f(x) = sqrt(x+1)
    g(x) = sqrt(x-1)

    f*g
    = sqrt(x+1) * sqrt(x-1)
    = sqrt[(x+1)(x-1)]
    = sqrt(x^2 - 1)

    f/g
    = sqrt(x+1) / sqrt(x-1)
    = sqrt[(x+1)(x-1)] / x-1

    thanks
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  2. #2
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    Hello, snypeshow!

    Looks good to me . . . Nice work!

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  3. #3
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    thanks!

    I have another question that i'm having trouble with.

    f(x) = 10^x
    g(x) = log x

    f(g(x)) = 10^logx
    g(f(x)) = log (10^x)

    how would i simplify f(g(x)) and g(f(x)) ?
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  4. #4
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    First of all, I assume that $\displaystyle log = lg$ which is the common logarithm.

    $\displaystyle f(x) = 10^x$
    $\displaystyle g(x) = log(x)$

    Now it is obvious that
    $\displaystyle (f \circ g)(x) = f(g(x)) = f(log(x)) = 10^{log(x)} = x$ and
    $\displaystyle (g \circ f)(x) = g(f(x)) = g(10^x) = log(10^x) = x$

    This is followed by the definition of logarithms:
    If $\displaystyle x = b^y$, then $\displaystyle y = log_b x$.

    When we simply make the substitution $\displaystyle x = b^y$ in $\displaystyle y = log_b x$, we get $\displaystyle y = log_b b^y$.
    We can also substitute $\displaystyle y = log_b x$ in If $\displaystyle x = b^y$ when we get $\displaystyle x = b^{log_b x}$.

    We also note that since $\displaystyle (f \circ g)(x) = x$ and $\displaystyle (g \circ f)(x) = x$, f and g are each other's inverse functions.

    By the way, do not forget about the range of definition when dealing with functions.
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