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Math Help - g/f(x)...

  1. #1
    Junior Member
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    Post g/f(x)...

    --->Question:
    f(x) = sqrt(4*x+5)
    g(x) = (x-2)/x

    find g/f(x) and it's domain



    --->My Answer:
    -find domain for both f(x) and g(x)
    f(x)-- sqrt(4*x+5) ---- under square root has to be greater than or = to zero
    4x+5 >= 0
    4x >= -5
    x >= -5/4

    g(x) -- the denominator cannot be zero
    x =! 0

    so for (g/f)(x), g has to be defined, f has to be defined and f can't be zero

    ((x-2)/x) * (1/(sqrt(4x+5)))
    x =! 0 & x > -5/4 [[[[no longer greater than or equal to 0]]]]]

    so domain in interval notation is
    (-5/4,0)U(0,infinity)




    DID I DO THIS RIGHT!?!?
    thanks for your help
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  2. #2
    Super Member
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    Quote Originally Posted by williamb View Post
    --->Question:
    f(x) = sqrt(4*x+5)
    g(x) = (x-2)/x

    find g/f(x) and it's domain



    --->My Answer:
    -find domain for both f(x) and g(x)
    f(x)-- sqrt(4*x+5) ---- under square root has to be greater than or = to zero
    4x+5 >= 0
    4x >= -5
    x >= -5/4

    g(x) -- the denominator cannot be zero
    x =! 0

    so for (g/f)(x), g has to be defined, f has to be defined and f can't be zero

    ((x-2)/x) * (1/(sqrt(4x+5)))
    x =! 0 & x > -5/4 [[[[no longer greater than or equal to 0]]]]]

    so domain in interval notation is
    (-5/4,0)U(0,infinity)




    DID I DO THIS RIGHT!?!?
    thanks for your help
    The function  \frac{g}{f}(x) = \frac{\frac{x-2}{x}}{\sqrt{4x+5}}

    Let's re-write that:

     \frac{g}{f}(x) = \frac{x-2}{x} \times \frac{1}{\sqrt{4x+5}}


      = \frac{x-2}{x\sqrt{4x+5}}

    Now, we need to consider which values this is defined for!

    The square root on the bottom must NOT be negative OR zero. In other words, it's positive.

     4x+5 >  0

     4x > -5

     x > \frac{-5}{4}

    The other x on the bottom cannot be zero, so  x \neq 0

    Hence the domain is  \bigg(\frac{-5}{4}, 0\bigg) \cup \bigg(0, \infty\bigg)
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