(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find the range for the function;

fg(x) and state its range where

[tex] f(x)= e^{2x} + 3[/tex] and [tex] g(x) = ln(x-1) [/tex]

3. The attempt at a solution

Putting g into f I reach;

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

[tex] fg(x) = (x-1)^2 + 3 [/tex] (2)

However the question is, can [tex] x = 1 [/tex], 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.

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# Homework Help: Domain and Range (simple)

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