If $g(x) = x^2 + 2x + 1$ and $(g \circ f)(x) = 4x^2 - 10x + 5$, then $f'(1)$ is equal to
Step-by-Step Solution
Key Concept: Domain of a quotient requires non-zero denominator and domain of square root requires non-negative radicand.
The domain requires $|x| - 1 \neq 0$ (so $|x| \neq 1$) and $x^2 + 4x + 2 \geq 0$. For Case 1 where $|x| - 1 < -1$ (i.e., $|x| < 0$, impossible) and Case 2 where $|x| - 1 > -1$ (i.e., $|x| > 0$, always true except $|x|=1$). Solving $x^2 + 4x + 2 \geq 0$ gives $x \in (-\infty, -3] \cup [-1, \infty)$. Excluding $x = \pm 1$ gives the final domain.
Correct Answer: $(-\infty, -3] \cup (-2, -1) \cup (2, \infty)$