题目

Problem

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

Given that

(a) find the value of $A$ .

(4)

Given also that

(b) find $f(x)$ , giving each term in simplest form.

(5)

解答

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先把 $f'(x)$ 写成指数形式，再求导得到 $f''(x)$ 。然后代入 $x=4$ 和 $f''(4)=0$ 求 $A$ 。

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Rewrite

\begin{align*} f'(x)=2x^{-\frac12}+Ax^{-2}+3. \end{align*}

Differentiate:

\begin{align*} f''(x) =&\,2\left(-\frac12\right)x^{-\frac32} -2Ax^{-3}\\ =&\,-x^{-\frac32}-2Ax^{-3}. \end{align*}

Since $f''(4)=0$ ,

\begin{align*} -4^{-\frac32}-2A(4^{-3})=&\,0. \end{align*}

Now

\begin{align*} 4^{-\frac32}=\frac18, \qquad 4^{-3}=\frac1{64}. \end{align*}

\begin{align*} -\frac18-\frac{2A}{64}=&\,0\\ -\frac18-\frac{A}{32}=&\,0\\ A=&\,-4. \end{align*}

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把 $A=-4$ 代回 $f'(x)$ ，积分得到 $f(x)$ 。再用 $f(12)=8\sqrt3$ 求常数。

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Using $A=-4$ ,

\begin{align*} f'(x)=2x^{-\frac12}-4x^{-2}+3. \end{align*}

Integrate:

\begin{align*} f(x) =&\,\int \left(2x^{-\frac12}-4x^{-2}+3\right)\,dx\\ =&\,4x^{\frac12}+4x^{-1}+3x+c\\ =&\,4\sqrt{x}+\frac4x+3x+c. \end{align*}

Use $f(12)=8\sqrt3$ :

\begin{align*} 8\sqrt3 =&\,4\sqrt{12}+\frac4{12}+3(12)+c\\ =&\,8\sqrt3+\frac13+36+c. \end{align*}

Hence

\begin{align*} c=-\frac{109}{3}. \end{align*}

Therefore

\begin{align*} f(x)=4\sqrt{x}+\frac4x+3x-\frac{109}{3}. \end{align*}