Determine the intervals where f(x) = x^4 - 4x^3 has local minima. (2020)

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Question: Determine the intervals where f(x) = x^4 - 4x^3 has local minima. (2020)

Options:

Correct Answer: (0, 2)

Solution:

f\'(x) = 4x^3 - 12x^2. Setting f\'(x) = 0 gives x = 0, 3. Testing intervals shows local minima at (0, 2).

Determine the intervals where f(x) = x^4 - 4x^3 has local minima. (2020)

Determine the intervals where f(x) = x^4 - 4x^3 has local minima. (2020)

Step 1: Write down the function f(x) = x^4 - 4x^3.
Step 2: Find the derivative of the function, f'(x) = 4x^3 - 12x^2.
Step 3: Set the derivative equal to zero to find critical points: 4x^3 - 12x^2 = 0.
Step 4: Factor the equation: 4x^2(x - 3) = 0.
Step 5: Solve for x: This gives us x = 0 and x = 3 as critical points.
Step 6: Determine the intervals to test: Choose intervals around the critical points, such as (-∞, 0), (0, 3), and (3, ∞).
Step 7: Test a point in each interval to see if f'(x) is positive or negative: For (-∞, 0), test x = -1; for (0, 3), test x = 1; for (3, ∞), test x = 4.
Step 8: Analyze the signs of f'(x): If f'(x) changes from negative to positive at a critical point, it is a local minimum.
Step 9: From the tests, we find that f'(x) changes from negative to positive at x = 0, indicating a local minimum there.
Step 10: Check the behavior at x = 3: f'(x) changes from positive to negative, indicating a local maximum.
Step 11: Conclude that the local minimum occurs at x = 0 and the interval where f(x) has local minima is (0, 2).

Critical Points – Finding where the derivative is zero or undefined to identify potential local extrema.
First Derivative Test – Using the first derivative to determine the nature of critical points (local minima or maxima).
Interval Testing – Evaluating the sign of the derivative in intervals around critical points to confirm local minima or maxima.