Notes on Second Derivative and Critical Points

  • Second Derivative and Critical Points

    • The function discussed is a second derivative:
    • f′′(x)=12−12xf''(x) = 12 - 12x
    • To find critical points, set the second derivative equal to zero:
    • 12−12x=012 - 12x = 0
    • Solving gives:
      • x=1x = 1
    • The second derivative equals zero at two points from the first derivative's perspective:
    • From the first derivative: critical points at:
      • x=0x = 0
      • x=2x = 2

  • First Derivative Test

    • Evaluate the intervals created by tested critical points:
    • Check intervals: ((-\infty, 0), (0, 2), (2, +\infty))
    • Choose test points for evaluation:
    • For Interval ((-\infty, 0)):
      • Test point: x=−1x = -1
      • Result: f′(−1)=−12−6=−18f'(-1) = -12 - 6 = -18 (negative)
    • For Interval ((0, 1)):
      • Test point: x=1x = 1
      • Result: f′(1)=12−6=6f'(1) = 12 - 6 = 6 (positive)
    • For Interval ((1, 3)):
      • Test point: x=3x = 3
      • Result: f′(3)=36−54=−18f'(3) = 36 - 54 = -18 (negative)
    • In conclusion:
    • Function is increasing on: ((0, 1))
    • Function is decreasing on: ((1, +\infty))

  • Second Derivative Test

    • To determine concavity: evaluate the second derivative around critical points:
    • For x=0x = 0:
      • Result: f′′(0)=12f''(0) = 12 (positive)
      • Concave up
    • For x=2x = 2:
      • Result: f′′(2)=12−24=−12f''(2) = 12 - 24 = -12 (negative)
      • Concave down
    • Conclusively:
    • Function is concave up in ((- ext{infinity}, 0)) and ((0, 2))
    • Function is concave down in ((2, +\text{infinity}))

  • Summary of Intervals

    • Increasing and concave up:
    • Interval: ((0, 1))
    • Decreasing and concave down:
    • Intervals: ((1, 2)) and ((2, +\infty))
    • Therefore, the final result indicates critical analysis of increasing and concave aspects of the function based on derivations and tests performed.