Examples of using the derivative tests for finding maximum/minimum

  • Problem 1: Find the maximum value of the function f(x) = -2x^2 + 4x + 1 .
  • Solution:
    • First, find the derivative f'(x) = -4x + 4 .
    • Set the derivative equal to zero to find critical points:
      -4x + 4 = 0 ? x = 1 .
    • To determine if this critical point is a maximum or minimum, find the second derivative f''(x) = -4 .
    • Since f''(x) < 0 , the function has a maximum at x = 1 .
    • The maximum value is f(1) = -2(1)^2 + 4(1) + 1 = 3 .

Problem 2: Find the minimum distance from the point (2, 3) to the line y = x + 1 .

Solution:

  1. Let the point on the line be (x, x + 1) .
  2. The distance D from (2, 3) to (x, x + 1) is given by:
    D = \sqrt{(x - 2)^2 + ((x + 1) - 3)^2}
  3. Simplify the distance formula:
    D = \sqrt{(x - 2)^2 + (x - 2)^2} = \sqrt{2(x - 2)^2} = \sqrt{2}|x - 2|
  4. To minimize D , minimize |x - 2| . The minimum value of |x - 2| is 0, which occurs at x = 2 .
  5. Substitute x = 2 back into the distance formula:
    D = \sqrt{2}(2 - 2) = 0
  6. Check if x = 2 is a minimum using derivatives:
    D^2 = 2(x - 2)^2
  7. Differentiate D^2 with respect to x :
    \frac{d(D^2)}{dx} = 4(x - 2)
    Set the derivative to 0:
    4(x - 2) = 0 \implies x = 2
  8. The second derivative test:
    \frac{d^2(D^2)}{dx^2} = 4
    Since \frac{d^2(D^2)}{dx^2} > 0 , x = 2 is indeed a minimum.

More examples:


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