题目

Problem

In this question you must show all stages of your working.

The curve $C$ has equation $y=f(x)$ , $x>0$ .

Given that

(a) find the exact value of $k$ .

(4)

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

(4)

解答

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题目给出的是 $f'(x)$ ，要使用 $f''(x)=0$ ，所以先对 $f'(x)$ 再求一次导。然后把点 $P$ 的 $x$ 坐标 $2$ 代入 $f''(x)=0$ ，解出 $k$ 。

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Rewrite $f'(x)$ as

\begin{align*} f'(x)=4x^{\frac32}+kx^{-2}. \end{align*}

Differentiate:

\begin{align*} f''(x) =&\,4\cdot\frac32x^{\frac12}-2kx^{-3}\\ =&\,6x^{\frac12}-2kx^{-3}. \end{align*}

Since $f''(x)=0$ at $P$ and $P$ has $x$ coordinate $2$ ,

\begin{align*} 0=&\,6(2)^{\frac12}-2k(2)^{-3}\\ 0=&\,6\sqrt2-\frac{k}{4}. \end{align*}

Hence

\begin{align*} \frac{k}{4}=&\,6\sqrt2\\ k=&\,24\sqrt2. \end{align*}

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先把 (a) 的 $k$ 代入 $f'(x)$ ，再积分得到 $f(x)$ 。最后用 $P(2,8\sqrt2)$ 求积分常数。

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Using $k=24\sqrt2$ ,

\begin{align*} f'(x)=4x^{\frac32}+24\sqrt2\,x^{-2}. \end{align*}

Integrate:

\begin{align*} f(x) =&\,4\cdot\frac{x^{\frac52}}{\frac52} +24\sqrt2\cdot\frac{x^{-1}}{-1}+c\\ =&\,\frac85x^{\frac52}-24\sqrt2\,x^{-1}+c\\ =&\,\frac85x^{\frac52}-\frac{24\sqrt2}{x}+c. \end{align*}

Use $P(2,8\sqrt2)$ :

\begin{align*} 8\sqrt2 =&\,\frac85(2)^{\frac52}-\frac{24\sqrt2}{2}+c\\ =&\,\frac85(4\sqrt2)-12\sqrt2+c\\ =&\,\frac{32\sqrt2}{5}-\frac{60\sqrt2}{5}+c\\ =&\,-\frac{28\sqrt2}{5}+c. \end{align*}

\begin{align*} c =&\,8\sqrt2+\frac{28\sqrt2}{5}\\ =&\,\frac{40\sqrt2}{5}+\frac{28\sqrt2}{5}\\ =&\,\frac{68\sqrt2}{5}. \end{align*}

Therefore

\begin{align*} f(x)=\frac85x^{\frac52}-\frac{24\sqrt2}{x}+\frac{68\sqrt2}{5}. \end{align*}