At which point does the function f(x) = -x^3 + 3x^2 + 4 have a local maximum? (2023)

1 question

1 item

Q: At which point does the function f(x) = -x^3 + 3x^2 + 4 have a local maximum? (2023)

Solution: Setting f'(x) = -3x^2 + 6x = 0 gives x = 0 and x = 2. f''(2) < 0 indicates a local maximum at (2, 5).

Steps: 10

Step 1: Write down the function f(x) = -x^3 + 3x^2 + 4.
Step 2: Find the first derivative f'(x) to determine where the function's slope is zero. The first derivative is f'(x) = -3x^2 + 6x.
Step 3: Set the first derivative equal to zero: -3x^2 + 6x = 0.
Step 4: Factor the equation: -3x(x - 2) = 0.
Step 5: Solve for x: This gives us x = 0 and x = 2.
Step 6: To determine if these points are local maxima or minima, find the second derivative f''(x). The second derivative is f''(x) = -6x + 6.
Step 7: Evaluate the second derivative at x = 2: f''(2) = -6(2) + 6 = -6, which is less than 0.
Step 8: Since f''(2) < 0, this indicates a local maximum at x = 2.
Step 9: Finally, find the value of the function at x = 2: f(2) = -2^3 + 3(2^2) + 4 = -8 + 12 + 4 = 8.
Step 10: Therefore, the local maximum occurs at the point (2, 8).