题目

Problem

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

Given that

(a) find the equation of the tangent to $C$ at $P$ . Write your answer in the form $y=mx+c$ , where $m$ and $c$ are constants to be found.

(3)

(b) Find, in simplest form, $f(x)$ .

(5)

解答

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切线斜率是 $f'(8)$ 。求出斜率后，用点 $P(8,2)$ 写直线方程。

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\begin{align*} f'(8) =&\,\frac{32}{3(8)^2}+3-2\sqrt[3]{8}\\ =&\,\frac{32}{192}+3-2(2)\\ =&\,\frac16-1\\ =&\,-\frac56. \end{align*}

Using $P(8,2)$ ,

\begin{align*} y-2=&\,-\frac56(x-8)\\ y=&\,-\frac56x+\frac{20}{3}+2\\ y=&\,-\frac56x+\frac{26}{3}. \end{align*}

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已知导函数，要求原函数，所以积分。积分后用 $P(8,2)$ 求常数。

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Write $f'(x)$ using indices:

\begin{align*} f'(x)=\frac{32}{3}x^{-2}+3-2x^{1/3}. \end{align*}

Integrate:

\begin{align*} f(x) =&\,\int\left(\frac{32}{3}x^{-2}+3-2x^{1/3}\right)\,dx\\ =&\,\frac{32}{3}\cdot\frac{x^{-1}}{-1} +3x-2\cdot\frac{x^{4/3}}{4/3} +c\\ =&\,-\frac{32}{3}x^{-1}+3x-\frac32x^{4/3}+c. \end{align*}

Use $P(8,2)$ :

\begin{align*} 2 =&\,-\frac{32}{3}(8)^{-1}+3(8)-\frac32(8)^{4/3}+c\\ =&\,-\frac43+24-\frac32(16)+c\\ =&\,-\frac43+c. \end{align*}

Thus

\begin{align*} c=\frac{10}{3}. \end{align*}

Therefore

\begin{align*} f(x)=-\frac{32}{3x}+3x-\frac32x^{4/3}+\frac{10}{3}. \end{align*}