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Determine the local maxima or minima of f(x) = -x^2 + 4x. (2019)
Determine the local maxima or minima of f(x) = -x^2 + 4x. (2019)
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Practice Questions
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Q1
Determine the local maxima or minima of f(x) = -x^2 + 4x. (2019)
Maxima at x=2
Minima at x=2
Maxima at x=4
Minima at x=4
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f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. Since f''(x) = -2 < 0, it is a maxima.
Questions & Step-by-step Solutions
1 item
Q
Q: Determine the local maxima or minima of f(x) = -x^2 + 4x. (2019)
Solution:
f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. Since f''(x) = -2 < 0, it is a maxima.
Steps: 7
Show Steps
Step 1: Write down the function f(x) = -x^2 + 4x.
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. Rearranging gives -2x = -4, so x = 2.
Step 5: To determine if this critical point is a maximum or minimum, find the second derivative f''(x). The second derivative is f''(x) = -2.
Step 6: Analyze the second derivative. Since f''(x) = -2 is less than 0, this indicates that the function has a local maximum at x = 2.
Step 7: To find the maximum value, substitute x = 2 back into the original function: f(2) = -2^2 + 4(2) = -4 + 8 = 4.
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