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# P2 Jan 03 Q8e. watch

1. Don't get this at all!

The question is:

f(x) = 3(2^-x) -1

Thanks!
2. f(x) = 3(2^-x) -1

g(x) = log2 x

fg(x) = f[g(x)]

= 3(2^-log2x) - 1
= [3/(2^log2x) ] - 1

2^log2 = 1 , so they basically cancell leaving just x

therefore = (3/x) - 1
3. fg(x) means apply g(x), then apply f(x). Thus, you put log(2)x into the first equation.

Since alogx = logx^a, this gives 3(2^log(2)[1/x]) - 1

= 3(1/x) - 1
= 3/x - 1

Are you sure it's supposed to be minus one on the end?

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