Problem
(C) $x=\frac{29 \sin 61^{\circ}}{\sin 79^{\circ}}$ (D) $x=\frac{21 \sin 40^{\circ}}{\sin 79^{\circ}}$ 9 Let $f(x)=x^{3}-\left(k^{2}-4\right) x+1$ For which of the values of $k$ below is $f(x)$ many-to-one? (A) $k=2$ (B) $k=-2$ (C) $k=1$ (D) $k=-4$
Solution
Step 1 ：Find the derivative of the function: \(f'(x) = 3x^2 - (k^2 - 4)\)
Step 2 ：Find the critical points: \(x = \pm\frac{\sqrt{3k^2 - 12}}{3}\)
Step 3 ：Check for which values of k there are multiple critical points: \(k = 1, -4\)
Step 4 ：Final Answer: \(\boxed{1}, \boxed{-4}\)
