题目

Problem

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

Solutions relying entirely on calculator technology are not acceptable.

(a) Find the equation of the tangent to the curve with equation

\begin{align*} y=\frac14x^3-8x^{-\frac12} \end{align*}

at the point $P(4,12)$ .

Give your answer in the form $ax+by+c=0$ where $a$ , $b$ and $c$ are integers.

(5)

The curve with equation $y=f(x)$ also passes through the point $P(4,12)$ .

Given that

\begin{align*} f'(x)=\frac14x^3-8x^{-\frac12}, \end{align*}

(b) find $f(x)$ giving the coefficients in simplest form.

(5)

解答

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切线斜率来自导数。先对曲线方程求导，再代入 $x=4$ ，最后用点 $P(4,12)$ 写切线方程。

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Differentiate:

\begin{align*} \frac{dy}{dx} =&\,\frac14(3x^2)-8\left(-\frac12x^{-\frac32}\right)\\ =&\,\frac34x^2+4x^{-\frac32}. \end{align*}

At $x=4$ ,

\begin{align*} \frac{dy}{dx} =&\,\frac34(4)^2+4(4)^{-\frac32}\\ =&\,12+4\cdot\frac18\\ =&\,\frac{25}{2}. \end{align*}

The tangent at $P(4,12)$ is

\begin{align*} y-12=&\,\frac{25}{2}(x-4)\\ 2y-24=&\,25x-100\\ 25x-2y-76=&\,0. \end{align*}

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这里给的是 $f'(x)$ ，所以要积分得到 $f(x)$ 。积分后用 $P(4,12)$ 求常数。

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\begin{align*} f(x) =&\,\int\left(\frac14x^3-8x^{-\frac12}\right)\,dx\\ =&\,\frac{1}{16}x^4-16x^{\frac12}+c. \end{align*}

Use $P(4,12)$ :

\begin{align*} 12 =&\,\frac{1}{16}(4)^4-16(4)^{\frac12}+c\\ =&\,16-32+c\\ =&\,-16+c. \end{align*}

\begin{align*} c=28. \end{align*}

Therefore

\begin{align*} f(x)=\frac{1}{16}x^4-16x^{\frac12}+28. \end{align*}