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

y=\frac{4x+1}{(x+3)^2}\qquad x\ne -3\qquad x\in\mathbb{R}

Use calculus to find the range of values of $x$ for which $y$ is increasing.

(6)

题目中文翻译

在本题中，你必须写出所有推导步骤。不接受依赖计算器技术的解法。

一条曲线的方程为

y=\frac{4x+1}{(x+3)^2}\qquad x\ne -3\qquad x\in\mathbb{R}

使用微积分求 $y$ 递增时 $x$ 的取值范围。

解答

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先把函数写成 $(4x+1)(x+3)^{-2}$ ，使用乘积法则求导并因式分解。判断递增区间时，不仅要找 $\dfrac{\mathrm{d}y}{\mathrm{d}x}=0$ 的点，还必须把原函数无定义的 $x=-3$ 作为分区边界，再逐段判断导数符号。

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Write

y=(4x+1)(x+3)^{-2}.

Differentiating,

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,4(x+3)^{-2} -2(4x+1)(x+3)^{-3}\\ =&\,(x+3)^{-3} \big(4(x+3)-2(4x+1)\big)\\ =&\,\frac{10-4x}{(x+3)^3}. \end{align*}

The derivative is zero when

10-4x=0,

x=\frac52.

The other boundary is $x=-3$ , where the function is undefined. A sign analysis gives

\begin{array}{c|ccc} x&(-\infty,-3)&(-3,\frac52)&(\frac52,\infty)\\ \hline 10-4x&+&+&-\\ (x+3)^3&-&+&+\\ \dfrac{\mathrm{d}y}{\mathrm{d}x}&-&+&- \end{array}

Therefore, $y$ is increasing when

\boxed{-3<x<\frac52}.

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也可以直接使用商法则。把导数分子因式分解成 $(x+3)(10-4x)$ 后，与分母约分；但约分不能让人忘记原函数在 $x=-3$ 处仍然无定义。

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Using the quotient rule,

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{ 4(x+3)^2-(4x+1)\,2(x+3)} {(x+3)^4}\\ =&\,\frac{ (x+3)\big(4(x+3)-2(4x+1)\big)} {(x+3)^4}\\ =&\,\frac{10-4x}{(x+3)^3}. \end{align*}

For the derivative to be positive, the numerator and denominator must have the same sign. This occurs only when

x>-3 \quad\text{and}\quad x<\frac52.

Hence

\boxed{-3<x<\frac52}.