Classifying stationary points

What does the derivative mean?

• If \frac{dy}{dx} > 0, the function is increasing (as x increases, y increases)
• If \frac{dy}{dx} < 0, the function is decreasing (as x increases, y decreases)
• If \frac{dy}{dx} = 0, the function is stationary

What does the second derivative mean?

If \frac{dy}{dx} = 0 at some point x = a, then we can use the second derivative to find the type of stationary point:

• If \frac{d^2y}{dx^2} > 0, the point is a local minimum.
• If \frac{d^2y}{dx^2} < 0, the point is a local maximum.
• If \frac{d^2y}{dx^2} = 0, it could be a point of inflection, but may not be. We can’t say for sure what type of stationary point it is.

Example: find the stationary points of y=x^3-3x^2+4 and classify each

• \frac{dy}{dx} = 3x^2 - 6x
• Set \frac{dy}{dx} = 0:
  • 3x^2 - 6x = 0
  • 3x(x - 2) = 0
  • x = 0 or x = 2
• Substitute back to find y-coordinates:
  • At x = 0: y = 0^3 - 3(0)^2 + 4 = 4 -> stationary point at (0, 4)
  • At x = 2: y = 2^3 - 3(2)^2 + 4 = 8 - 12 + 4 = 0 -> stationary point at (2, 0)
• Now we find the second derivative:
  • \frac{d^2y}{dx^2} = 6x - 6
• Evaluate the second derivative at each stationary point:
  • At x = 0: \frac{d^2y}{dx^2} = 6(0) - 6 = -6 (less than 0) -> Local maximum at (0, 4)
  • At x = 2: \frac{d^2y}{dx^2} = 6(2) - 6 = 6 (greater than 0) -> Local minimum at (2, 0)
• Answer: Local maximum at (0, 4) and local minimum at (2, 0)

Example: find the stationary points of y=x^4-4x^3+6x^2 and classify each

• \frac{dy}{dx} = 4x^3 - 12x^2 + 12x
• Set \frac{dy}{dx} = 0:
  • 4x^3 - 12x^2 + 12x = 0
  • 4x(x^2 - 3x + 3) = 0
  • x = 0 or x = \frac{3 \pm \sqrt{3^2 - 4\times1\times3}}{2\times1}
  • x = 0 or x = \frac{3 \pm \sqrt{-3}}{2}
  • x = 0 (the other two solutions are complex numbers)
• Substitute back to find y-coordinate:
  • At x = 0: y = 0^4 - 4(0)^3 + 6(0)^2 = 0 -> stationary point at (0, 0)
• Now we find the second derivative:
  • \frac{d^2y}{dx^2} = 12x^2 - 24x + 12
• Evaluate the second derivative at the stationary point:
  • At x = 0: \frac{d^2y}{dx^2} = 12(0)^2 - 24(0) + 12 = 12 (greater than 0) -> Local minimum at (0, 0)
• Answer: Local minimum at (0, 0)

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