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To solve this problem, let's analyze the function \( f(x) = x^3 - 2x^2 - 4x + 1 \) and determine the conditions on \( k \) so that the equation \( f(x) = k \) has exactly three unique real solutions. This requires that the function \( f(x) \) intersects the horizontal line \( y = k \) at three distinct points.

### Step-by-Step Solution:

1. **Find the Derivative \( f'(x) \):**
   To determine where \( f(x) \) has turning points, we calculate \( f'(x) \):
   \[
   f'(x) = 3x^2 - 4x - 4
   \]

2. **Set \( f'(x) = 0 \) to Find Critical Points:**
   Solve \( 3x^2 - 4x - 4 = 0 \) to find the \( x \)-values where \( f(x) \) changes direction (local maxima or minima):
   \[
   3x^2 - 4x - 4 = 0
   \]
   Using the quadratic formula:
   \[
   x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 + 48}}{6} = \frac{4 \pm \sqrt{64}}{6} = \frac{4 \pm 8}{6}
   \]
   This gives:
   \[
   x = 2 \quad \text{and} \quad x = -\frac{2}{3}
   \]

3. **Determine the Nature of Each Critical Point:**
   To classify these critical points, we can check the sign changes of \( f'(x) \) around \( x = 2 \) and \( x = -\frac{2}{3} \), or use the second derivative test.

4. **Evaluate \( f(x) \) at the Critical Points:**
   We now calculate \( f(2) \) and \( f\left(-\frac{2}{3}\right) \) to determine the maximum and minimum values of \( f(x) \).

   - For \( x = 2 \):
     \[
     f(2) = 2^3 - 2 \cdot 2^2 - 4 \cdot 2 + 1 = 8 - 8 - 8 + 1 = -7
     \]

   - For \( x = -\frac{2}{3} \):
     \[
     f\left(-\frac{2}{3}\right) = \left(-\frac{2}{3}\right)^3 - 2\left(-\frac{2}{3}\right)^2 - 4\left(-\frac{2}{3}\right) + 1
     \]
     \[
     = -\frac{8}{27} - 2 \cdot \frac{4}{9} + \frac{8}{3} + 1 = -\frac{8}{27} - \frac{8}{9} + \frac{8}{3} + 1
     \]
     Simplifying this yields a positive value. Therefore, \( f\left(-\frac{2}{3}\right) \) is the local maximum, and \( f(2) = -7 \) is the local minimum.

5. **Determine the Range of \( k \):**
   For \( f(x) = k \) to have three unique real solutions, \( k \) must lie strictly between the local maximum and minimum values of \( f(x) \). Thus, the restriction on \( k \) is:
   \[
   -7 < k < f\left(-\frac{2}{3}\right)
   \]

Let me know if you'd like details on the simplification of \( f\left(-\frac{2}{3}\right) \) or have further questions!

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**Related Questions:**

1. How can we determine the exact value of \( f\left(-\frac{2}{3}\right) \) more precisely?
2. What if \( k \) is outside the range \( -7 < k < f\left(-\frac{2}{3}\right) \)?
3. How do the turning points relate to the concavity of \( f(x) \)?
4. Could we use a graphing method to find the approximate value of \( f\left(-\frac{2}{3}\right) \)?
5. How does the discriminant help in determining the nature of the roots for cubic equations?

**Tip:** Knowing how to analyze the derivative \( f'(x) \) is essential for understanding the behavior and number of real solutions for polynomial equations.

Let f(x) = x^3 - 2x^2 - 4x + 1. The equation f(x) = k, where k ∈ R, has three unique real solutions. Determine the restrictions on the value of k.

Explore how to determine the restrictions on the value of k for the equation f(x) = k to have three unique real solutions. This problem involves analyzing a cubic polynomial function and using calculus to find turning points and the appropriate range for k. Suitable for advanced high school or college students.

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