SolveMathAI

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

Generated by SolveMath AI and kept here so the next person asking gets it instantly. AI can make mistakes, so check anything that matters.

Solve your own

Same solver, same steps, no account.

Enter to solve · Shift+Enter for a new line

  • Free
  • No account
  • Every step shown
  • Arithmetic to calculus

More solved problems

Calculus solver →
Find the maximum value of f(x) = -x^2 + 4x + 1Find the average value of f(x) = x^2 on [0, 3]Determine whether the series sum of 1/n^2 convergesFind the Taylor series of e^x about x = 0 up to the x^3 termProve that the sum of the first n odd numbers equals n squared, using induction