Discussion
Let \(f(x) = x^{2} - 10x\) and \(g(x) = e^{x} + 5\) for \(x \in \mathbb{R}\). Then, for all \(x\), \(g(2x) - (f \circ g)(x) =\)
Answer: A
The value is 30.
Compute each part.
\[ \begin{aligned} g(2x) &= e^{2x} + 5 \\ f(g(x)) &= (e^{x} + 5)^{2} - 10(e^{x} + 5) \\ &= e^{2x} + 10e^{x} + 25 - 10e^{x} - 50 \\ &= e^{2x} - 25 \end{aligned} \]
\[ g(2x) - (f \circ g)(x) = (e^{2x} + 5) - (e^{2x} - 25) = 30 \]
The \(e^{2x}\) terms cancel, so the result is the same constant for every \(x\). Getting \(-30\) means the subtraction was done the other way round, and 20 or \(-20\) come from mishandling the constant 5.
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