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Practice Question
A-LevelHard

A curve has equation y = x³ - 3x² + 4. Find the coordinates of all stationary points and determine their nature.

Hint - Step 1 of 3

Stationary points occur where the gradient equals zero. Start by finding dy/dx.

Differentiate: if y = x³ − 3x² + 4, then dy/dx = 3x² − 6x
Full Solution
Step 1: dy/dx = 3x² − 6x = 3x(x − 2)
Step 2: Set dy/dx = 0 → x = 0 or x = 2
Step 3: y(0) = 4, y(2) = 0. Points: (0, 4) and (2, 0)
Step 4: d²y/dx² = 6x − 6. At x=0: −6 (maximum). At x=2: +6 (minimum)
AI Coach - Reviewing Your Answer

Good attempt - you correctly differentiated. The key error was not setting dy/dx = 0 before solving. Stationary points require the gradient to be zero, not just finding dy/dx.

Set 3x² − 6x = 0, factorise to 3x(x − 2) = 0, giving x = 0 and x = 2.
Always use the second derivative d²y/dx² to classify each point as a maximum or minimum.
Similar Questions
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Classify turning points of y = x⁴ − 4x³ + 6Hard
3
A curve passes through (1,5). Given dy/dx = 3x² − 2, find y.Med

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