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IsabelleTreuille
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IsabelleTreuille
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Bunuel
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IsabelleTreuille
Thank you so much for replying!

In the second way to solve:
f(f(−π)) = f(π − (−π)) = f(π + π) = f(2π) = π − 2π = −π.

I don't understand at all how we get from
this: f(f(−π))
to this: f(π − (−π))
to this: f(π + π)

Also, I don't usually see explanations turning unknown function questions into classic 'functions' with f(x) - do you know if there is a reason why this is?

Check function questions in our Special Questions Directory:

Operations/functions defining algebraic/arithmetic expressions
Symbols Representing Arithmetic Operation
Rounding Functions
Various Functions

Hope it helps.

P.S. Please read and follow: rules-for-posting-please-read-this-before-posting-133935.html Thank you.
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wainsdaylion
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"this: f(f(−π))
to this: f(π − (−π))"

Ignore the wrapper function (the extra f() ) for a moment...

The function is f ( x ) = π - x, so
f ( π ) = π - x
f ( π ) = π - ( - π )

Is that any clearer? If so...subtracting a negative is the same as adding.
π - ( - π ) = π + π

Why is there a question using an unusual function format instead of the standard f()? I do not question the why of GMAT! But I suspect it's testing your mental flexibility, that you grasp we're just dealing in symbols that we can arbitrarily define and that x* has nothing to do with the multiplication or exponential functions but is simply a function we're newly defining.
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Given:
x* = π - x

Find:
((−π)*)*

Step 1:
(−π)* = π - (−π) = π + π = 2π

Step 2:
(2π)* = π - 2π = −π

Final Answer:
−π
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