The maximum slope of the tangents to the curve \(y(x) = -x^3 + 3x^2 + 9x - 30\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 12
The slope of the tangent at any point is \(\frac{dy}{dx}\):
\(m(x) = -3x^2 + 6x + 9\)
We need the maximum of m(x), so differentiate again:
\(m'(x) = -6x + 6 = 0\)
\(x = 1\)
\(m''(x) = -6 \lt 0\), so m is maximum at x = 1.
\(m(1) = -3 + 6 + 9 = 12\)
NCERT Exemplar:
Application of Derivatives
