\[f'(x) = \frac{d}{dx}(x^3 - 3x + 2) = 3x^2 - 3\] Set the derivative equal to zero to find the critical points:\[3x^2 - 3 = 0\]\[3(x^2 - 1) = 0\]\[x^2 = 1\]\[x = \pm 1\]The critical points are \(x = -1\) and \(x = 1\). Since the interval is \([0, 2]\), we only consider the critical point \(x = 1\), as \(x = -1\) is outside the interval.2. Evaluate Function at Critical Points and Endpoints
📍The absolute maximum value of a continuous function on a closed interval must occur at a critical point within the interval or at the endpoints of the interval.
We evaluate \(f(x)\) at \(x=1\), \(x=0\), and \(x=2\).At the critical point \(x=1\):\[f(1) = (1)^3 - 3(1) + 2 = 1 - 3 + 2 = \mathbf{0}\]At the left endpoint \(x=0\):\[f(0) = (0)^3 - 3(0) + 2 = 0 - 0 + 2 = \mathbf{2}\]At the right endpoint \(x=2\):\[f(2) = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = \mathbf{4}\]
The largest value is 4.Therefore, the absolute maximum value of the function \(f(x) = x^3 - 3x + 2\) in the interval \([0, 2]\) is \(\mathbf{4}\).
AI Tutor Explanation
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Step-by-Step Solution
To find the absolute maximum value of the function \( f(x) = x^3 - 3x + 2 \) in the interval [0, 2], we need to follow these steps:
1. **Find the critical points:**
* Take the derivative of \( f(x) \):
\[ f'(x) = 3x^2 - 3 \]
* Set \( f'(x) = 0 \) and solve for \( x \):
\[ 3x^2 - 3 = 0 \]
\[ 3x^2 = 3 \]
\[ x^2 = 1 \]
\[ x = \pm 1 \]
* The critical points are \( x = 1 \) and \( x = -1 \). Since we are only considering the interval [0, 2], we only consider \( x = 1 \).
2. **Evaluate the function at the critical points and endpoints:**
* Evaluate \( f(x) \) at \( x = 0 \), \( x = 1 \), and \( x = 2 \):
* \( f(0) = (0)^3 - 3(0) + 2 = 2 \)
* \( f(1) = (1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0 \)
* \( f(2) = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = 4 \)
3. **Determine the absolute maximum value:**
* Comparing the values \( f(0) = 2 \), \( f(1) = 0 \), and \( f(2) = 4 \), the absolute maximum value is 4.
Correct Answer: 4
Pedagogical Audit
Bloom's Analysis:
This is an APPLY question because it requires students to apply the concepts of derivatives and finding critical points to determine the absolute maximum value of a function within a given interval.
Knowledge Dimension:PROCEDURAL
Justification:The question requires a step-by-step procedure to find the derivative, critical points, and evaluate the function at those points and endpoints.
Syllabus Audit:
In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding and application of calculus concepts taught in the textbook.
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