题目

Problem

Figure 1 shows a sketch of a curve with equation $y=f(x)$ where

f(x)=\frac{2x+3}{\sqrt{4x-1}}\qquad x>\frac{1}{4}

(a) Find, in simplest form, $f'(x)$ .

(4)

(b) Hence find the range of $f$ .

(3)

题目中文翻译

图 1 给出了曲线 $y=f(x)$ 的草图，其中

f(x)=\frac{2x+3}{\sqrt{4x-1}}\qquad x>\frac{1}{4}

(a) 求 $f'(x)$ 的最简形式。

(b) 由此求 $f$ 的值域。

解答

Write

f(x)=\frac{2x+3}{\sqrt{4x-1}}

f(x)=(2x+3)(4x-1)^{-\frac12}

Differentiate using the product rule:

\begin{aligned} f'(x) &=2(4x-1)^{-\frac12} +(2x+3)\left(-\frac12\right)(4x-1)^{-\frac32}(4)\\ &=2(4x-1)^{-\frac12}-2(2x+3)(4x-1)^{-\frac32} \end{aligned}

Factor out

2(4x-1)^{-\frac32}

to get

\begin{aligned} f'(x) &=2(4x-1)^{-\frac32}\left[(4x-1)-(2x+3)\right]\\ &=2(4x-1)^{-\frac32}(2x-4)\\ &=4(x-2)(4x-1)^{-\frac32} \end{aligned}

Therefore

\boxed{f'(x)=\frac{4x-8}{(4x-1)^{\frac32}}}

For stationary points,

f'(x)=0

Using part (a),

\frac{4x-8}{(4x-1)^{\frac32}}=0

Since

x>\frac14

the denominator is positive. So

4x-8=0

Therefore

x=2

Now find the corresponding value of $f(x)$ :

\begin{aligned} f(2) &=\frac{2(2)+3}{\sqrt{4(2)-1}}\\ &=\frac{7}{\sqrt7}\\ &=\sqrt7 \end{aligned}

The curve decreases until $x=2$ and then increases after $x=2$ , so this is the minimum value.

Therefore the range of $f$ is

\boxed{f(x)\geq \sqrt7}

or equivalently

\boxed{[\sqrt7,\infty)}