Domain and range of composite function

[Gwyddel]

Homework Statement

The function f has domain (-∞, ∞) and is defined by f(x) = 3e^2x.

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 3e^2x 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) = 3e^2ln4x

= 3e^ln(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.

 

[jedishrfu]

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?

 

[HallsofIvy]

Homework Statement

The function f has domain (-∞, ∞) and is defined by f(x) = 3e^2x.

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 3e^2x 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) = 3e^2ln4x

= 3e^ln(4x)2​

= 3ln16x²​

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

= 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.

 

[Gwyddel]

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

Oh! That was very silly of me.

3e^ln(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?

 

[eumyang]

Oh! That was very silly of me.

3e^ln(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?

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.

 

[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).

 

[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.

 

[Gwyddel]

the domain of f is (-oo, 0) U (0, oo).

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

 

[eumyang]

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

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

 

[checkitagain]

Edit

The domain of g is (0, oo), and the domain of
f is (-oo, oo). This is corrected now.

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.

 

[Gwyddel]

Thank you all for your valuable help!