A real-valued differentiable function y = f(x) is implicitly defined by the relation y² + y - x = 2 for y in (-inf, -1/2). If y = g(x) is the inverse function of y = f(x), then |g'(-2)| equals:

Correct answer: 3

Solution

The inverse g satisfies the equation x = g(x)² + g(x) - 2; equivalently g(x) = x² + x - 2 (for x < -1/2), so g'(x) = 2x + 1 and |g'(-2)| = |2(-2)+1| = |-3| = 3.

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