题目

Problem

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

Given that

(a) find the value of $a$ .

(3)

(b) Hence find $f(x)$ .

(4)

解答

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已知 $f'(x)$ ，所以先再求一次导数得到 $f''(x)$ 。然后把 $x=27$ 代入，并令 $f''(27)=0$ 。

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\begin{align*} f'(x)=&\,ax-12x^{1/3},\\ f''(x)=&\,a-12\cdot\frac13x^{-2/3}\\ =&\,a-4x^{-2/3}. \end{align*}

Since $f''(x)=0$ when $x=27$ ,

\begin{align*} 0=&\,a-4(27)^{-2/3}. \end{align*}

Now

\begin{align*} 27^{1/3}=3, \quad\text{so}\quad 27^{-2/3}=\frac19. \end{align*}

Hence

\begin{align*} 0=&\,a-\frac49\\ a=&\,\frac49. \end{align*}

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把 (a) 的 $a=\frac49$ 代回 $f'(x)$ ，然后积分求 $f(x)$ 。最后用点 $(1,-8)$ 求积分常数。

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Using $a=\frac49$ ,

\begin{align*} f'(x)=\frac49x-12x^{1/3}. \end{align*}

Integrate:

\begin{align*} f(x) =&\,\int\left(\frac49x-12x^{1/3}\right)\,dx\\ =&\,\frac49\cdot\frac{x^2}{2} -12\cdot\frac{x^{4/3}}{4/3} +c\\ =&\,\frac29x^2-9x^{4/3}+c. \end{align*}

The curve passes through $(1,-8)$ , so

\begin{align*} -8 =&\,\frac29(1)^2-9(1)^{4/3}+c\\ =&\,\frac29-9+c\\ =&\,-\frac{79}{9}+c. \end{align*}

Thus

\begin{align*} c=\frac79. \end{align*}

Therefore

\begin{align*} f(x)=\frac29x^2-9x^{4/3}+\frac79. \end{align*}