Discussion
If \(\phi(x) = x^{2}\) and \(\psi(x) = 2^{x}\), then \(\psi(\phi(x))\) is
Answer: A
\(\psi(\phi(x)) = 2^{x^{2}}\).
Composition means applying \(\phi\) first and then \(\psi\) to the result. Since \(\phi(x) = x^{2}\), substitute \(x^{2}\) for \(x\) in \(\psi(x) = 2^{x}\):
\[ \psi(\phi(x)) = \psi(x^{2}) = 2^{x^{2}} \]
\(x^{2}\) is \(\phi(x)\) alone. \(2^{2x}\) would be \(\psi(2x)\), which doubles \(x\) instead of squaring it. \(x^{2x}\) mixes the two functions without composing them. Note also that the order matters: \(\phi(\psi(x)) = (2^{x})^{2} = 2^{2x}\), which is a different function.
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