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Re: x# = x2 + 3x x§ = x2 + 2x [#permalink]
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Explanation

Whenever you have a function within another function, you have to first calculate the value of the function on the inside, and then plug that value into the function on the outside.

Plug in to test the values. If you plug in 2 for x, then, in Quantity A, 2$ = \(2^2 + 2(2) = 8\)and 8# =\(8^2 + 3(8) = 88\); in Quantity B, 2# = \(2^2 + 3(2) = 10\) and 10$ = \(10^2 + 2(10) = 120\); eliminate choices (A) and (C).

However, if you plug in 0 for x, then both quantities are 0; eliminate choice (B), and you’re left with choice (D), the correct answer.
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Re: x# = x2 + 3x x§ = x2 + 2x [#permalink]
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Well, this is a tricky question.. because (XS)# is not a function like f(g(x)) that you have to solve the inner function and use the value in the encompassing function as @ Sandy noted above. The figure (XS)# and (x#)S are both multiplication of function, not a function within a function. So the answer could be "C"... because both resolve to X^3+5X^2+6X.
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Re: x# = x2 + 3x x§ = x2 + 2x [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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