The tangent to the curve \(y = 4x^{3} + kx^{2} - 6x + 4\) at the point P(1, m) is parallel to the x- axis, where k and m are constants. Determine the coordinates of P.
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Correct Answer: Option C
Explanation:
\(y = 4x^{3} + kx^{2} - 6x + 4\)
\(\frac{\mathrm d y}{\mathrm d x} = 12x^{2} + 2kx - 6\)
At P(1, m)
\(\frac{\mathrm d y}{\mathrm d x} = 12 + 2k - 6 = 0\) (parallel to the x- axis)
\(6 + 2k = 0 \implies k = -3\)
\(P(1, m) \implies m = 4(1^{3}) - 3(1^{2}) - 6(1) + 4)
= -1
P = (1, -1)
\(y = 4x^{3} + kx^{2} - 6x + 4\)
\(\frac{\mathrm d y}{\mathrm d x} = 12x^{2} + 2kx - 6\)
At P(1, m)
\(\frac{\mathrm d y}{\mathrm d x} = 12 + 2k - 6 = 0\) (parallel to the x- axis)
\(6 + 2k = 0 \implies k = -3\)
\(P(1, m) \implies m = 4(1^{3}) - 3(1^{2}) - 6(1) + 4)
= -1
P = (1, -1)