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Math Help - Functions

  1. #1
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    Functions

    Functions f and g are defined for x E R by

    f:x --->e^x
    g:x--->2x-3

    (i) Solve the equation fg(x)=7

    Function h is defined as gf
    (ii)Express h in terms of x and state its range
    (iii)Express h inverse in terms of x.
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  2. #2
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    Re: Functions

    (i)x=(ln7+3)/2...its easy. take ln both sides.
    (ii)h(x)=2e^x-3
    its minimum value will be when the term 2e^x is minimum since 3 is constant ..2e^x can attain a minimum zero(just attain) at x=-infinity.And its maximum value will be infinity.therefore range is (-3,infinity)
    (iii)for finding h inverse take log again..
    Hope it helps....
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  3. #3
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    Re: Functions

    Note that:

    f \circ g(x) = f(g(x))

    So, in this case, that means:

    f \circ g(x) = e^{g(x)} = e^{2x-3} = (e^{2x})(e^{-3}) = \frac{(e^x)^2}{e^3} = 7

    This leads to:

    e^x = \sqrt{7e^3}, and taking logs, we get:

    x = \log(\sqrt{7e^3}) = \log((\sqrt{7})(e^{\frac{3}{2}}))

     = \log(\sqrt{7}) + \log(e^{\frac{3}{2}}) = \log(\sqrt{7}) + \frac{3}{2} = \frac{\log(7) + 3}{2}

    Similarly,

    g \circ f(x) = 2(e^x) - 3.

    Since e^x ranges from (0,∞), it's clear that 2(e^x) also ranges from (0,∞) and thus the range of g \circ f is (-3,∞).

    Consequently, h^{-1} = f^{-1} \circ g^{-1} is only defined on (-3,∞). Can you figure out what it is, in terms of x?
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  4. #4
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    Re: Functions

    i solved it maybe it is : (ln(x+3)/2)=h-1(x)
    here domain of h-1(x) is (-3,infinity) same as the range ..
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