critical points of x^3-6x^2+8 at interval [1,6]

To find the critical points of the function \( f(x) = x^3 - 6x^2 + 8 \) in the interval \([1, 6]\), we’ll follow these steps:

### Step 1: Find the Derivative
Calculate \( f'(x) \) to find the critical points.
\[
f(x) = x^3 - 6x^2 + 8
\]
Taking the derivative:
\[
f'(x) = 3x^2 - 12x
\]

### Step 2: Set the Derivative Equal to Zero
To find critical points, set \( f'(x) = 0 \):
\[
3x^2 - 12x = 0
\]
Factor out \( 3x \):
\[
3x(x - 4) = 0
\]
This gives us two solutions:
\[
x = 0 \quad \text{and} \quad x = 4
\]

### Step 3: Check if the Critical Points Are in the Interval \([1, 6]\)
The interval of interest is \([1, 6]\), so we only consider \( x = 4 \) (since \( x = 0 \) is outside the interval).

### Step 4: Evaluate the Function at the Endpoints and at \( x = 4 \)
To find the values of \( f(x) \) at the endpoints and at \( x = 4 \), we calculate:
\[
f(1) = 1^3 - 6 \cdot 1^2 + 8 = 1 - 6 + 8 = 3
\]
\[
f(4) = 4^3 - 6 \cdot 4^2 + 8 = 64 - 96 + 8 = -24
\]
\[
f(6) = 6^3 - 6 \cdot 6^2 + 8 = 216 - 216 + 8 = 8
\]

### Summary of Critical Points
The critical point in the interval \([1, 6]\) is:
- \( x = 4 \), with \( f(4) = -24 \).

### Function Values at the Endpoints
- At \( x = 1 \), \( f(1) = 3 \).
- At \( x = 6 \), \( f(6) = 8 \).

### Conclusion
The critical point within the interval \([1, 6]\) is at \( x = 4 \), with \( f(4) = -24 \).

Would you like further details or have any questions about this process?

Here are some follow-up questions for deeper understanding:
1. How does finding the second derivative help determine the nature of a critical point?
2. What is the significance of evaluating function values at endpoints when examining intervals?
3. How can one confirm if a critical point is a maximum or minimum within a given interval?
4. Can a function have critical points that aren’t maxima or minima? If so, why?
5. How does the Fundamental Theorem of Calculus relate to finding critical points in intervals?

**Tip:** Always check endpoints when determining absolute extrema in a closed interval, as the highest or lowest points may occur there.

Critical points of x^3 - 6x^2 + 8 at interval [1,6]

Learn how to find the critical points of the function f(x) = x^3 - 6x^2 + 8 within the interval [1,6]. This detailed solution covers differentiation, setting the derivative equal to zero to find critical points, and evaluating function values at endpoints. Perfect for high school students in calculus classes.

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