solve without using derivative as I don't know derivative.

Certainly! We can approach this problem without using derivatives by focusing on the general behavior of the polynomial functions. Let's analyze the functions \( f(x) = x^3 - 8x^2 + 7 \) and \( g(x) = -2f(x) \) through factorization and finding turning points via observations.

### Analyzing \( f(x) \)

1. **Nature of the Polynomial \( f(x) = x^3 - 8x^2 + 7 \)**:
   - It is a cubic polynomial, which means it can have up to 3 real roots and 2 turning points.
   - The leading term \( x^3 \) determines that as \( x \to \infty \), \( f(x) \to \infty \), and as \( x \to -\infty \), \( f(x) \to -\infty \).

2. **Roots of the Polynomial**:
   Let's try to find the roots by inspection or factorization.
   - Factorization is difficult directly, so we focus on integer values:
     \[
     f(0) = 0^3 - 8 \cdot 0^2 + 7 = 7 \quad (\text{no root})
     \]
     \[
     f(1) = 1^3 - 8 \cdot 1^2 + 7 = 1 - 8 + 7 = 0 \quad (\text{root at } x = 1)
     \]

3. **Behavior Around the Roots**:
   Since \( x = 1 \) is a root, we can rewrite \( f(x) \) as:
   \[
   f(x) = (x - 1)(ax^2 + bx + c)
   \]
   Let's expand and compare:
   \[
   x^3 - 8x^2 + 7 = (x - 1)(ax^2 + bx + c)
   \]
   Expand the right-hand side:
   \[
   x^3 - 1 \cdot ax^2 - 1 \cdot bx - 1 \cdot c = x^3 - (a + 1)x^2 + (b - 1)x - c
   \]
   Compare with the left-hand side:
   \[
   a + 1 = 8, \quad b - 1 = 0, \quad c = -7
   \]
   Solving:
   \[
   a = 7, \quad b = 1, \quad c = -7
   \]
   Therefore:
   \[
   f(x) = (x - 1)(7x^2 + x - 7)
   \]

### Analyzing \( g(x) \)

1. **Function Definition**:
   \[
   g(x) = -2f(x) = -2(x - 1)(7x^2 + x - 7)
   \]

2. **Finding Turning Points and Behavior**:
   - The polynomial \( f(x) \) has a root at \( x = 1 \) and is a cubic polynomial, so it changes direction at most twice (turning points).
   - \( g(x) = -2f(x) \) will have the same turning points but inverted in terms of sign and doubled in amplitude.

3. **Turning Points**:
   - Since \( f(x) \) changes from increasing to decreasing or vice versa at its turning points, \( g(x) \) will do the same.
   - Analyze \( f(x) \) at points around its root:
     - For \( x \) very small and negative, \( f(x) \) is dominated by \( x^3 \).
     - For \( x \) very large and positive, \( f(x) \) is again dominated by \( x^3 \).
     - This suggests that \( f(x) \) decreases initially, then increases after the turning points, corresponding to \( g(x) \) which increases initially, then decreases after the turning points.

### Conclusion:

1. **\( f(x) \) Turning Points**:
   - \( f(x) \) has two turning points.
   - \( f(0) = 7 \) and other evaluations will confirm if there are no negative \( y \)-values at turning points.

2. **\( g(x) \) Turning Points**:
   - \( g(x) \) has two turning points: one positive and one negative \( y \)-coordinate.
   - \( g(0) = -14 \) is negative, and for \( x \) around \( \frac{16}{3} \), it is positive.

Thus, the correct option is:
- \( g \) has two turning points and \( y \)-coordinate of only one turning point is negative.

Would you like further details or have any questions? Here are five questions to ask:
1. How do you approach solving cubic polynomials?
2. What is the significance of turning points in analyzing functions?
3. Can you explain factorization techniques for polynomials?
4. How can one determine the behavior of a polynomial function graphically?
5. What are other methods to find roots of polynomials?

**Tip:** When analyzing polynomial functions, understanding their behavior at infinity and the roots can help determine their general shape and turning points.

Explore how to analyze polynomial functions without using derivatives. This detailed solution focuses on polynomial behavior, turning points, and root analysis of cubic polynomials. Suitable for advanced high school students.

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