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# Composite function help please. :) Watch

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1. Hello there,

Cold someone tell me how I would work out fg where f(x) = 3e^2x and g(x)=ln4x. I know how to form a composite function i'm just stuck on what to do as there is and e and ln in the equation. Thanks!
2. The same way as you usually would. What do you get when you do it? Post your working. Also, is that exp(2x) or exp^2(x)?
3. (Original post by little_wizard123)
The same way as you usually would. What do you get when you do it? Post your working. Also, is that exp(2x) or exp^2(x)?
Hi, it's exp^2x, sorry.

My working is:-

(3)(e^(2)(ln4x)) = fg. Is this correct? Is this all I can do? Thanks!
4. You can use a log rule to simplify it. And then use the fact that ln(x) and e^x are inverses of each other.
5. (Original post by apo1324)
Hi, it's exp^2x, sorry.

My working is:-

(3)(e^(2)(ln4x)) = fg. Is this correct? Is this all I can do? Thanks!
If you mean , then that's right.

You can simplify it more though. Note that
6. (Original post by Daniel Freedman)
If you mean , then that's right.

You can simplify it more though. Note that
Is the answer just 12x? Thanks. Also, if the domain of f was (-infinite,infinite), and the domain of g was (0,infinite):-

Would the domain of fg be [0,infinite) and the range of fg be [0,infinite)?? I don't know if this is correct.
7. (Original post by apo1324)
Is the answer just 12x? Thanks. Also, if the domain of f was (-infinite,infinite), and the domain of g was (0,infinite):-

Would the domain of fg be [0,infinite) and the range of fg be [0,infinite)?? I don't know if this is correct.
Nope. You can't use exp (ln(x)) = x if there is a number in front of the log. You need to use the rule that a*ln(x) = ln(x^a) first.
8. (Original post by little_wizard123)
Nope. You can't use exp (ln(x)) = x if there is a number in front of the log. You need to use the rule that a*ln(x) = ln(x^a) first.
Is it this:-

3e^ln4x^2?
9. (Original post by apo1324)
Is it this:-

3e^ln4x^2?
(4x)^2 which is 16x^2.

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Updated: December 7, 2010
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