Let f(x)=x3−6x2+3f(x)=x3−6x2+3

Find the critical numbers of ff: 

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 (Separate multiple answers by commas.)

Determine the intervals on which ff is increasing and decreasing. *Help entering intervals*ff is increasing on: 

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ff is decreasing on: 

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Use the First Derivative Test to determine whether each critical point is a relative maximum, minimum, or neither. (Separate multiple answers by commas, if there is no answer enter “none”.)Relative maxima occur at x=x=

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Relative minima occur at x=x=

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To analyze the function \( f(x) = x^3 - 6x^2 + 3 \), we need to find its critical numbers, determine the intervals on which it is increasing or decreasing, and use the First Derivative Test to classify the critical points.

### Step 1: Finding Critical Numbers
Critical numbers occur where the derivative \( f'(x) \) is zero or undefined.

1. Compute the first derivative \( f'(x) \):
   \[
   f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 3) = 3x^2 - 12x
   \]
2. Set the derivative equal to zero and solve for \( x \):
   \[
   3x^2 - 12x = 0
   \]
   Factor out \( 3x \):
   \[
   3x(x - 4) = 0
   \]
   So the critical numbers are:
   \[
   x = 0, \quad x = 4
   \]

### Step 2: Determining Intervals of Increase and Decrease
To determine where \( f(x) \) is increasing or decreasing, we analyze the sign of \( f'(x) \) on the intervals determined by the critical points.

1. Choose test points in the intervals \((-\infty, 0)\), \((0, 4)\), and \((4, \infty)\).

   - For \( x \in (-\infty, 0) \):
     \[
     f'(-1) = 3(-1)^2 - 12(-1) = 3 + 12 = 15 \quad (\text{positive})
     \]
   - For \( x \in (0, 4) \):
     \[
     f'(2) = 3(2)^2 - 12(2) = 12 - 24 = -12 \quad (\text{negative})
     \]
   - For \( x \in (4, \infty) \):
     \[
     f'(5) = 3(5)^2 - 12(5) = 75 - 60 = 15 \quad (\text{positive})
     \]

Therefore, \( f(x) \) is:
- Increasing on \( (-\infty, 0) \cup (4, \infty) \)
- Decreasing on \( (0, 4) \)

### Step 3: First Derivative Test for Relative Extrema
Use the First Derivative Test to classify the critical points \( x = 0 \) and \( x = 4 \).

1. **At \( x = 0 \)**:
   - \( f'(x) \) changes from positive to negative.
   - Hence, \( x = 0 \) is a relative maximum.

2. **At \( x = 4 \)**:
   - \( f'(x) \) changes from negative to positive.
   - Hence, \( x = 4 \) is a relative minimum.

### Summary
- **Critical numbers**: \( x = 0, 4 \)
- **Increasing intervals**: \( (-\infty, 0) \cup (4, \infty) \)
- **Decreasing intervals**: \( (0, 4) \)
- **Relative maxima**: \( x = 0 \)
- **Relative minima**: \( x = 4 \)

Would you like more details or have any questions?

Here are some related questions:
1. How do we find the second derivative of \( f(x) \) and use it to confirm concavity?
2. What is the behavior of \( f(x) \) as \( x \) approaches \(\pm\infty\)?
3. Can we identify any points of inflection for \( f(x) \)?
4. How do the relative extrema of \( f(x) \) relate to the overall graph of the function?
5. What are the coordinates of the relative extrema points for \( f(x) \)?
6. How can we sketch the graph of \( f(x) \) using the critical points and intervals of increase/decrease?
7. What is the absolute maximum and minimum of \( f(x) \) on a closed interval, e.g., \([0, 5]\)?
8. How does the graph of \( f(x) = x^3 - 6x^2 + 3 \) compare to other cubic functions?

**Tip**: When analyzing a function, always check for critical points and intervals of increase/decrease to understand its behavior thoroughly.

Learn how to find critical numbers, intervals of increase and decrease, and classify relative extrema for the function f(x) = x^3 - 6x^2 + 3 using the First Derivative Test. Suitable for advanced high school students and calculus learners.

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