题目

Problem

In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

A curve has equation

\begin{align*} y=x(x+3)(x-2) \end{align*}

(a) Find, in simplest form, $\dfrac{\mathrm{d}y}{\mathrm{d}x}$

(3)

(b) Hence find the range of values for $x$ such that

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}\geqslant2 \end{align*}

(4)

解答

展开

先把三次式展开成标准多项式，再逐项求导。

展开

\begin{align*} y=&\,x(x+3)(x-2)\\ =&\,x(x^2+x-6)\\ =&\,x^3+x^2-6x. \end{align*}

Differentiate term by term:

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =3x^2+2x-6. \end{align*}

展开

把 (a) 的导函数代入不等式，整理成二次不等式。由于二次项系数为正，满足大于等于 $0$ 的区间在两个根的外侧。

展开

Using part (a),

\begin{align*} 3x^2+2x-6&\geqslant2\\ 3x^2+2x-8&\geqslant0. \end{align*}

Factorise:

\begin{align*} 3x^2+2x-8 =&\,(3x-4)(x+2). \end{align*}

\begin{align*} (3x-4)(x+2)&\geqslant0. \end{align*}

The critical values are

\begin{align*} x=-2,\qquad x=\frac43. \end{align*}

Since the quadratic is positive or zero outside the roots,

\begin{align*} x\leqslant-2 \quad\text{or}\quad x\geqslant\frac43. \end{align*}