Question

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) = $}

• A. 30
• B. 20
• C. -30
• D. -20
• E. 15

Hint:

Notice that \( f(x) = (x-5)^2 - 25 \). Substituting \( g(x) = e^x + 5 \) makes \( f(g(x)) = ((e^x+5)-5)^2 - 25 = e^{2x} - 25 \). This "completing the square" method can save significant time.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We must evaluate two separate expressions: \(g(2x)\) and the composition \(f(g(x))\).
Then, we subtract the second from the first.
Step 2: Key Formula or Approach:
Substitute \(2x\) into \(g(x)\).
Substitute \(g(x)\) into \(f(x)\) using \((A + B)^2 = A^2 + 2AB + B^2\).
Step 3: Detailed Explanation:
Find \(g(2x)\):
\[ g(2x) = e^{2x} + 5 \] Find \(f(g(x))\):
\[ f(g(x)) = f(e^x + 5) = (e^x + 5)^2 - 10(e^x + 5) \] Expand the terms:
\[ = (e^{2x} + 10e^x + 25) - 10e^x - 50 \] \[ = e^{2x} - 25 \] Subtract the two results:
\[ g(2x) - f(g(x)) = (e^{2x} + 5) - (e^{2x} - 25) \] \[ = e^{2x} + 5 - e^{2x} + 25 = 30 \] Step 4: Final Answer:
The result is 30.