Domain and Range of fg(x) for f(x)= e^{2x} + 3 and g(x) = ln(x-1)

[Chewy0087]

Homework Statement

Find the range for the function;

fg(x) and state its range where

$$f(x)= e^{2x} + 3$$ and $$g(x) = ln(x-1)$$

The Attempt at a Solution

Putting g into f I reach;

$$fg(x) = e^{2ln(x-1)}+3$$ (1) - here is where I have the problem, now, I know that it simplifies down to

$$fg(x) = (x-1)^2 + 3$$ (2)

However the question is, can $$x = 1$$, it's obvious that for the function g, x > 1, or it's undefined, however when you put that into fg(x) the ln goes away so clearly in equation (2) fg >= 3, whereas with (1) fg > 3 (because x > 1). So basically I was wondering, it's obvious that g, f, & fg are linked, however is there a fundamental link between their domain/ranges? Or is it considered a new separate function?

I'd appreciate some guidance, this was on an exam the other day.

 

[Mark44]

(f o g)(x) = f(g(x)) = e^{2ln(x - 1)} + 3 = e^ln(x-1)2 + 3.

Simplify the exponential term using the fact that e^ln(u) = u, for u > 0.

On reading your post more carefully, I think you know this.

The composite function f o g is tied to the two functions that make it up, but is different from both. You need to look at the domain of g, which will have a bearing on the domain of f o g. The

 

[Chewy0087]

Hmmm so fg(x) is not a function of x in its own right? it's dependent on the functions that make it up? That doesn't really make sense to me...=/

As for the domain of g, I believe it's g > 0, but how does that help me?

 

[Mark44]

Hmmm so fg(x) is not a function of x in its own right? it's dependent on the functions that make it up? That doesn't really make sense to me...=/

Yes f o g is a function in its own right. Note that when you write fg, it's usually interpreted to mean the product of two functions. IOW (fg)(x) = f(x)*g(x). What you're working with is the composition of two functions, which is why I have been writing the letter o between them. f o g (x) = f(g(x)).

As for the domain of g, I believe it's g > 0, but how does that help me?

No that doesn't make any sense. g is the name of the function. For g(x) = ln(x - 1), the domain is {x| x > 1}. The input to the ln function has to be positive, so x - 1 > 0, so x > 1.

 

[Chewy0087]

Yes f o g is a function in its own right. Note that when you write fg, it's usually interpreted to mean the product of two functions. IOW (fg)(x) = f(x)*g(x). What you're working with is the composition of two functions, which is why I have been writing the letter o between them. f o g (x) = f(g(x)).

No that doesn't make any sense. g is the name of the function. For g(x) = ln(x - 1), the domain is {x| x > 1}. The input to the ln function has to be positive, so x - 1 > 0, so x > 1.

Ah yeah, sorry, i did mean f(g(x)), my bad.

So sorry for the confusion >.< but would the range therefore of f(g(x)) be >= 3? Sorry to be blunt but my teacher says it's just > 3 and i don't believe him >.<

 

[Mark44]

The range of f o g is {y | y > 3}. For the range to include 3, it would have to be true that e^{ln(x - 1)2} is 0, which doesn't happen for any real number x.

Think about it: e^{ln(x - 1)2} is identically equal to (x - 1)², for all x > 1. If x = 1, (x - 1)² = 0, but e^{ln(x - 1)2} is undefined.

Also, x <= 1 is not in the domain of f o g, for the reason that x <= 1 is not in the domain of ln(x - 1).