题目

Problem

The function $f$ is defined by

f(x)=\frac{49x}{x^2+x-12}+\frac{7x}{x+4}\qquad x>3

(a) Show that

f(x)=\frac{7x}{x-3}\qquad x>3

(3)

(b) Hence find $f'(x)$ giving your answer in simplest form.

(2)

题目中文翻译

函数 $f$ 定义为

f(x)=\frac{49x}{x^2+x-12}+\frac{7x}{x+4}\qquad x>3

(a) 证明

f(x)=\frac{7x}{x-3}\qquad x>3

(b) 由此求 $f'(x)$ ，并将答案化为最简形式。

解答

First factorise the denominator:

x^2+x-12=(x+4)(x-3)

f(x)=\frac{49x}{(x+4)(x-3)}+\frac{7x}{x+4}

Use the common denominator $(x+4)(x-3)$ :

\begin{aligned} f(x) &=\frac{49x}{(x+4)(x-3)} +\frac{7x(x-3)}{(x+4)(x-3)} \\ &=\frac{49x+7x(x-3)}{(x+4)(x-3)} \\ &=\frac{49x+7x^2-21x}{(x+4)(x-3)} \\ &=\frac{7x^2+28x}{(x+4)(x-3)} \\ &=\frac{7x(x+4)}{(x+4)(x-3)} \end{aligned}

Since $x>3$ , we have $x\ne -4$ , so we can cancel $x+4$ :

f(x)=\frac{7x}{x-3}

as required.

Using part (a),

f(x)=\frac{7x}{x-3}

Differentiate using the quotient rule:

\begin{aligned} f'(x) &=\frac{(x-3)(7)-7x(1)}{(x-3)^2} \\ &=\frac{7x-21-7x}{(x-3)^2} \\ &=-\frac{21}{(x-3)^2} \end{aligned}

Therefore

\boxed{f'(x)=-\frac{21}{(x-3)^2}}