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Homework Help: Domain and range of composite function

  1. Jan 4, 2012 #1
    1. The problem statement, all variables and given/known data
    The function f has domain (-∞, ∞) and is defined by f(x) = 3e2x.

    The function g has domain (0, ∞) and is defined by g(x) = ln 4x.

    (a) Write down the domain and range of f∘g.
    (b) Solve the equation (f∘g)(x) = 12

    2. The attempt at a solution


    Is it correct to think that the domain of f∘g will be all those x in the domain of g which produce g(x) in the domain of f? The domain of f is ℝ, so there's no restrictions on the g(x), so I think the domain of f∘g is the same as the domain of g, which is (0, ∞).

    Taking the limit of 3e2x as x→-∞ gives me 0, and x→∞ gives ∞. I got that the range of f is (0, ∞). I, uh, think the range of g is (-∞, ∞).

    I notice that the range of g and the domain of f are the same: (-∞, ∞). I therefore conclude that the range of f∘g will be the same as the range of f.

    Domain of f∘g: (0, ∞)
    Range of f∘g: (0, ∞)

    f∘g(x) = 3e2ln4x
    = 3eln(4x)2
    = 3ln16x2
    = 6ln16x​

    (6ln16x)/6 = 12/6
    ln16x = 2
    16x = e2
    x = e2/16 = 0.4618... ≈ 0.5

    I really don't know if any of this is right, and these questions always make me scratch my head. Especially any question on the domain and range of a composite function.
  2. jcsd
  3. Jan 4, 2012 #2


    Staff: Mentor

    first look at your functions and determine the ranges too.

    the range of one function will be the possible domain of values for the other.

    so say the domain of the inner function is -1 to 1 and the range of the inner function is -10 to 10 but the domain of the outer function 0 to 100 then what is the actual domain of values for the composite?

    I think it would be 0 to 1 right because the -1 to 0 is clipped off by the limit of 0 to 100 constraint of the outer.

    does that help without giving away the answer?
  4. Jan 4, 2012 #3


    User Avatar
    Science Advisor

    What happened to the "e"? ex and ln(x) are inverse functions so eln((4x)^2)= (4x)2= 16x2.

  5. Jan 4, 2012 #4
    Oh! That was very silly of me.

    3eln(4x)2 = 3(16x2) = 48x2

    48x2 = 12
    x2 = 12/48 = 1/2

    x = 1/√2, x = -1/√2

    Exclude x = -1/√2 because it is outside the domain of fg?
  6. Jan 4, 2012 #5


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    Homework Helper

    Checked the bolded above. Also, I assume you mean f ° g, not fg. The former is a function composition, and the latter is a multiplication of functions. They are not the same thing.
  7. Jan 4, 2012 #6
    Ah, okay, so correcting my silly arithmetic error.

    x2 = 12/48 = 1/4
    x = plusorminus sqrt(1/4) = 1/ (plusorminus sqrt(4))
    x= 1/2, -1/2 (which is not accepted because it is not on the interval (0, ∞), the domain of f ° g).
  8. Jan 5, 2012 #7
    The domain of g is (0, oo), and the domain of f is (-oo, 0) U (0, oo).

    The domain of 2ln(4x) is (0, oo), as opposed to the domain of ln[(4x)^2],

    which is (-oo, 0) U (0, oo).

    Then the domain of the composite function, f(g(x)) = 3e^{2ln(4x)}, is (0, oo).

    The graph of the composite function is the right half portion of a parabola.
  9. Jan 5, 2012 #8
    Why is 0 not in the domain of 3exp(2x) ?
  10. Jan 5, 2012 #9


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    Homework Helper

    0 is in the domain of f. checkitagain is wrong.
  11. Jan 5, 2012 #10

  12. Jan 5, 2012 #11
    Thank you all for your valuable help!
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