Find the local maximum of f(x) = -x^3 + 3x^2 + 4. (2020)

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Q: Find the local maximum of f(x) = -x^3 + 3x^2 + 4. (2020)

Solution: Set f'(x) = 0 to find critical points. The local maximum occurs at x = 2. f(2) = 5.

Steps: 12

Step 1: Write down the function f(x) = -x^3 + 3x^2 + 4.
Step 2: Find the derivative of the function, f'(x). This tells us the slope of the function.
Step 3: Calculate f'(x) = -3x^2 + 6x.
Step 4: Set the derivative equal to zero to find critical points: -3x^2 + 6x = 0.
Step 5: Factor the equation: -3x(x - 2) = 0.
Step 6: Solve for x: This gives us x = 0 and x = 2 as critical points.
Step 7: To find out if these points are local maxima or minima, we can use the second derivative test.
Step 8: Find the second derivative, f''(x) = -6x + 6.
Step 9: Evaluate the second derivative at the critical points: f''(2) = -6(2) + 6 = -6 (which is less than 0).
Step 10: Since f''(2) < 0, x = 2 is a local maximum.
Step 11: Calculate the value of the function at the local maximum: f(2) = -2^3 + 3(2^2) + 4 = -8 + 12 + 4 = 8.
Step 12: Therefore, the local maximum occurs at x = 2 and the maximum value is 8.