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If f(x) = x^3 - 3x^2 + 4, then the local maxima and minima occur at which of the

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Question: If f(x) = x^3 - 3x^2 + 4, then the local maxima and minima occur at which of the following points?

Options:

  1. (0, 4)
  2. (1, 2)
  3. (2, 2)
  4. (3, 4)

Correct Answer: (1, 2)

Solution:

To find local maxima and minima, we first find f\'(x) = 3x^2 - 6x. Setting f\'(x) = 0 gives x(3x - 6) = 0, so x = 0 or x = 2. Evaluating f(1) = 2 shows it is a local minimum.

If f(x) = x^3 - 3x^2 + 4, then the local maxima and minima occur at which of the

Practice Questions

Q1
If f(x) = x^3 - 3x^2 + 4, then the local maxima and minima occur at which of the following points?
  1. (0, 4)
  2. (1, 2)
  3. (2, 2)
  4. (3, 4)

Questions & Step-by-Step Solutions

If f(x) = x^3 - 3x^2 + 4, then the local maxima and minima occur at which of the following points?
  • Step 1: Start with the function f(x) = x^3 - 3x^2 + 4.
  • Step 2: Find the derivative of the function, f'(x). The derivative 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 two solutions, x = 0 and x = 2.
  • Step 7: To determine if these points are local maxima or minima, we need to evaluate the function at these points.
  • Step 8: Calculate f(0) = 0^3 - 3(0^2) + 4 = 4.
  • Step 9: Calculate f(2) = 2^3 - 3(2^2) + 4 = 8 - 12 + 4 = 0.
  • Step 10: To find if these points are maxima or minima, we can check the value of the function at a point between them, like f(1).
  • Step 11: Calculate f(1) = 1^3 - 3(1^2) + 4 = 1 - 3 + 4 = 2.
  • Step 12: Since f(1) = 2 is less than both f(0) = 4 and f(2) = 0, x = 1 is a local minimum.
  • Step 13: Therefore, the local minima occurs at (1, 2) and the local maxima occurs at (0, 4) and (2, 0).
  • Finding Local Extrema – The question tests the understanding of how to find local maxima and minima using the first derivative test.
  • Critical Points – Identifying critical points by setting the first derivative equal to zero is a key concept in determining local extrema.
  • Evaluating Function Values – Evaluating the function at critical points to determine whether they are local maxima or minima is essential.
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