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Find the critical points of f(x) = x^3 - 3x^2

Worked through step by step. More calculus.

Answer
(0,0) and (2,−4)\boxed{(0, 0) \text{ and } (2, -4)}

Checking the result, the derivative 3x2−6x3x^2 - 6x is indeed zero at both x=0x = 0 and x=2x = 2.

Problem: Find the critical points of f(x)=x3−3x2f(x) = x^3 - 3x^2.

Steps

  1. Find the first derivative
    f′(x)=3x2−6xf'(x) = 3x^2 - 6x
  2. Set the derivative equal to zero
    3x2−6x=03x^2 - 6x = 0
  3. Factor the equation
    3x(x−2)=03x(x - 2) = 0
  4. Solve for xx
    x=0orx=2x = 0 \quad \text{or} \quad x = 2
  5. Find the corresponding yy-values by evaluating f(x)f(x)
    f(0)=03−3(0)2=0f(0) = 0^3 - 3(0)^2 = 0
    f(2)=23−3(2)2=8−12=−4f(2) = 2^3 - 3(2)^2 = 8 - 12 = -4

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