Question 16
The equation of a curve is such that $\dfrac{d^2y}{dx^2} = -\dfrac{24}{x^3}$. It is given that the curve has a stationary point at $(-2, 19)$.
**(b)** Find the $x$-coordinate of the other stationary point of the curve, and determine the nature of this stationary point.
(In part (a) you found $\dfrac{dy}{dx} = \dfrac{12}{x^2} - 3$.)
**(b)** Find the $x$-coordinate of the other stationary point of the curve, and determine the nature of this stationary point.
(In part (a) you found $\dfrac{dy}{dx} = \dfrac{12}{x^2} - 3$.)
