Find the stationary point of the curve \(y = 3x^{2} - 2x^{3}\).
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option C
Explanation:
\(y = 3x^{2} - 2x^{3}\)
\(\frac{\mathrm d y}{\mathrm d x} = 6x - 6x^{2} = 0 \implies 6x(1 - x) = 0\)
\(x = 0, 1\), when x = 0, y = 0.
when x = 1, y = 1.
The stationary point is (1, 1)
\(y = 3x^{2} - 2x^{3}\)
\(\frac{\mathrm d y}{\mathrm d x} = 6x - 6x^{2} = 0 \implies 6x(1 - x) = 0\)
\(x = 0, 1\), when x = 0, y = 0.
when x = 1, y = 1.
The stationary point is (1, 1)