4.
Determine the coordinates of the local maximum and the local minimum of
𝑦=𝑥3−6𝑥2+9𝑥 (4)
5. Draw a neat labelled sketch graph of 𝑓(𝑥)=−𝑥3+4𝑥2−4𝑥 (4)
6.
Find the equation of the tangent to the graph of 𝑓(𝑥)=+3𝑥3−2𝑥2+4𝑥−1 at the point where x = - 1. (4)
7.
Determine the coordinates of the point(s) on the graph of 𝑦=2𝑥3−5𝑥2−2𝑥+10 where the slope of the graph is equal to - 1 (4)

Let's address each of these questions step-by-step:

### 4. Local Maximum and Minimum of \( y = x^3 - 6x^2 + 9x \)

To find the local maxima and minima, we first find the derivative of the function and then set it to zero to find the critical points.

1. **Derivative:**
   \[
   y' = \frac{d}{dx}(x^3 - 6x^2 + 9x) = 3x^2 - 12x + 9
   \]

2. **Set the derivative to zero:**
   \[
   3x^2 - 12x + 9 = 0
   \]

3. **Solve for \( x \):**
   \[
   x^2 - 4x + 3 = 0
   \]
   Factorizing:
   \[
   (x - 3)(x - 1) = 0
   \]
   So, \( x = 3 \) or \( x = 1 \).

4. **Second derivative test:**
   - Calculate the second derivative \( y'' \):
     \[
     y'' = 6x - 12
     \]
   - Substitute \( x = 1 \) and \( x = 3 \) to determine the concavity:
     - For \( x = 1 \): \( y''(1) = 6(1) - 12 = -6 \) (negative, so local maximum)
     - For \( x = 3 \): \( y''(3) = 6(3) - 12 = 6 \) (positive, so local minimum)

5. **Calculate the corresponding \( y \) values:**
   - At \( x = 1 \): \( y = 1^3 - 6(1)^2 + 9(1) = 4 \)
   - At \( x = 3 \): \( y = 3^3 - 6(3)^2 + 9(3) = 0 \)

**Coordinates of Local Maxima and Minima:**
- Local Maximum: \( (1, 4) \)
- Local Minimum: \( (3, 0) \)

### 5. Sketch of \( f(x) = -x^3 + 4x^2 - 4x \)

1. **Find critical points:**
   - \( f'(x) = -3x^2 + 8x - 4 = 0 \)
   - Factorizing gives \( (x - 2)(x - \frac{2}{3}) = 0 \), so \( x = 2 \) and \( x = \frac{2}{3} \).

2. **Find second derivative to determine concavity:**
   - \( f''(x) = -6x + 8 \)
   - At \( x = 2 \): \( f''(2) = -4 \) (local maximum)
   - At \( x = \frac{2}{3} \): \( f''(\frac{2}{3}) = 4 \) (local minimum)

3. **Calculate \( f(x) \) at critical points and intercepts:**
   - At \( x = 0 \), \( f(0) = 0 \) (y-intercept)
   - At \( x = 2 \), \( f(2) = 0 \) (x-intercept)

The graph will have a local maximum at \( (2, 0) \) and a local minimum at \( (\frac{2}{3}, \frac{4}{27}) \).

### 6. Equation of Tangent Line at \( x = -1 \) for \( f(x) = 3x^3 - 2x^2 + 4x - 1 \)

1. **Find the derivative \( f'(x) \):**
   \[
   f'(x) = 9x^2 - 4x + 4
   \]
2. **Evaluate \( f'(-1) \):**
   \[
   f'(-1) = 9(-1)^2 - 4(-1) + 4 = 17
   \]
3. **Find \( f(-1) \):**
   \[
   f(-1) = 3(-1)^3 - 2(-1)^2 + 4(-1) - 1 = -10
   \]
4. **Tangent line equation:**
   \[
   y - (-10) = 17(x + 1)
   \]
   Simplifying:
   \[
   y = 17x + 7
   \]

### 7. Points on \( y = 2x^3 - 5x^2 - 2x + 10 \) with slope -1

1. **Find the derivative \( y' = 6x^2 - 10x - 2 \).**
2. **Set \( y' = -1 \):**
   \[
   6x^2 - 10x - 2 = -1 \implies 6x^2 - 10x - 1 = 0
   \]
3. **Solve the quadratic equation using the quadratic formula.**

These steps will help you determine the required solutions. Let me know if you need further clarification on any specific part!

### Related Questions:
1. How do you identify the nature of critical points using the second derivative?
2. What is the general approach for sketching the graph of a cubic function?
3. How do you find the equation of a tangent line at a given point?
4. What are the applications of finding the slope of a curve?
5. How do you solve a cubic equation?

#### Tip:
Always check your derivative calculations by revisiting the function’s behavior near the critical points to confirm maxima or minima.


Explore calculus problems including finding local maxima and minima, tangent lines, and slopes. This page provides detailed solutions to calculus questions involving derivatives and critical points. Suitable for undergraduate students and enthusiasts of calculus.

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