Maximum slope of the curve y=-x^3+3x^2+9x-30 is

The correct answer is Option (3) → 12

The slope of the curve is \( \frac{dy}{dx} \), and its maximum is found by differentiating the slope again.

    \( \displaystyle m=\frac{dy}{dx}=-3x^2+6x+9 \)

    \( \displaystyle \frac{dm}{dx}=-6x+6=0 \)

    \( x=1 \)

Since \( \frac{d^2m}{dx^2}=-6\lt 0 \), the slope is maximum at x = 1.

    \( m_{\max}=-3+6+9=12 \)

Shortcut: The slope \( -3x^2+6x+9 \) is a quadratic with a negative leading coefficient, so its greatest value is at the vertex:

    \( \displaystyle x=-\frac{b}{2a}=-\frac{6}{2(-3)}=1 \)

Putting x = 1 in the slope gives 12.

NCERT Exemplar:
Application of Derivatives