# Domain and range of composite function

1. Jan 4, 2012

### Gwyddel

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

(a)

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, ∞)

(b)
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. Jan 4, 2012

### 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?

3. Jan 4, 2012

### HallsofIvy

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

4. Jan 4, 2012

### Gwyddel

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?

5. Jan 4, 2012

### eumyang

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.

6. Jan 4, 2012

### Gwyddel

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).

7. Jan 5, 2012

### checkitagain

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.

8. Jan 5, 2012

### Gwyddel

Why is 0 not in the domain of 3exp(2x) ?

9. Jan 5, 2012

### eumyang

0 is in the domain of f. checkitagain is wrong.

10. Jan 5, 2012

### checkitagain

Edit

11. Jan 5, 2012

### Gwyddel

Thank you all for your valuable help!