Find the local maxima of f(x) = -x^3 + 3x^2 + 1. (2020)

₹0.0

Question: Find the local maxima of f(x) = -x^3 + 3x^2 + 1. (2020)

Options:

Correct Answer: (1, 3)

Exam Year: 2020

Solution:

f\'(x) = -3x^2 + 6x. Setting f\'(x) = 0 gives x(3x - 6) = 0, so x = 0 or x = 2. f(2) = 5.

Find the local maxima of f(x) = -x^3 + 3x^2 + 1. (2020)

Find the local maxima of f(x) = -x^3 + 3x^2 + 1. (2020)

Step 1: Write down the function we want to analyze: f(x) = -x^3 + 3x^2 + 1.
Step 2: Find the derivative of the function, which tells us the slope: f'(x) = -3x^2 + 6x.
Step 3: Set the derivative equal to zero to find critical points: -3x^2 + 6x = 0.
Step 4: Factor the equation: -3x(x - 2) = 0.
Step 5: Solve for x: This gives us two solutions: x = 0 and x = 2.
Step 6: To find the local maxima, we need to evaluate the function at these critical points: f(0) and f(2).
Step 7: Calculate f(0): f(0) = -0^3 + 3(0)^2 + 1 = 1.
Step 8: Calculate f(2): f(2) = -2^3 + 3(2)^2 + 1 = -8 + 12 + 1 = 5.
Step 9: Compare the values: f(0) = 1 and f(2) = 5. The highest value is at x = 2.
Step 10: Therefore, the local maximum of the function is at x = 2, and the maximum value is 5.

Finding Local Maxima – This involves taking the derivative of a function, setting it to zero to find critical points, and then using the second derivative test or evaluating the function to determine if these points are maxima or minima.
Critical Points – Identifying where the first derivative is zero or undefined to find potential local maxima or minima.
Second Derivative Test – Using the second derivative to determine the concavity at critical points to classify them as local maxima, minima, or points of inflection.