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Determine the local maxima and minima of f(x) = x^2 - 4x + 3.
Determine the local maxima and minima of f(x) = x^2 - 4x + 3.
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Practice Questions
1 question
Q1
Determine the local maxima and minima of f(x) = x^2 - 4x + 3.
Maxima at x=2
Minima at x=2
Maxima at x=1
Minima at x=1
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f'(x) = 2x - 4. Setting f'(x) = 0 gives x = 2. f''(x) = 2 > 0 indicates a local minimum at x = 2.
Questions & Step-by-step Solutions
1 item
Q
Q: Determine the local maxima and minima of f(x) = x^2 - 4x + 3.
Solution:
f'(x) = 2x - 4. Setting f'(x) = 0 gives x = 2. f''(x) = 2 > 0 indicates a local minimum at x = 2.
Steps: 7
Show Steps
Step 1: Write down the function f(x) = x^2 - 4x + 3.
Step 2: Find the first derivative f'(x) to determine the slope of the function. The first derivative is f'(x) = 2x - 4.
Step 3: Set the first derivative equal to zero to find critical points: 2x - 4 = 0.
Step 4: Solve for x: 2x = 4, so x = 2.
Step 5: Find the second derivative f''(x) to determine the concavity of the function. The second derivative is f''(x) = 2.
Step 6: Analyze the second derivative: since f''(x) = 2 is greater than 0, this indicates that the function is concave up at x = 2.
Step 7: Conclude that there is a local minimum at x = 2.
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