题目

Problem

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

The point $P\left(4,\dfrac{32}{3}\right)$ lies on the curve.

Given that

\begin{align*} f''(x)=\frac4{\sqrt{x}}-3 \end{align*}

and $f'(x)=5$ at $P$ , find

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

(2)

(b) $f(x)$ .

(8)

解答

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在点 $P$ 处， $f'(x)=5$ ，所以切线斜率是 $5$ 。直接用点斜式。

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Using $P\left(4,\frac{32}{3}\right)$ and gradient $5$ ,

\begin{align*} y-\frac{32}{3}=&\,5(x-4)\\ y=&\,5x-\frac{28}{3}. \end{align*}

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由二阶导数积分一次得到 $f'(x)$ ，用 $f'(4)=5$ 求第一个常数；再积分得到 $f(x)$ ，用点 $P$ 求第二个常数。

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Write

\begin{align*} f''(x)=4x^{-1/2}-3. \end{align*}

Integrate once:

\begin{align*} f'(x) =&\,4\cdot\frac{x^{1/2}}{1/2}-3x+k\\ =&\,8x^{1/2}-3x+k. \end{align*}

Use $f'(4)=5$ :

\begin{align*} 5=&\,8(4)^{1/2}-3(4)+k\\ =&\,16-12+k\\ =&\,4+k. \end{align*}

\begin{align*} k=1. \end{align*}

Thus

\begin{align*} f'(x)=8x^{1/2}-3x+1. \end{align*}

Integrate again:

\begin{align*} f(x) =&\,8\cdot\frac{x^{3/2}}{3/2} -\frac32x^2+x+d\\ =&\,\frac{16}{3}x^{3/2}-\frac32x^2+x+d. \end{align*}

Use $f(4)=\frac{32}{3}$ :

\begin{align*} \frac{32}{3} =&\,\frac{16}{3}(4)^{3/2}-\frac32(4)^2+4+d\\ =&\,\frac{128}{3}-24+4+d\\ =&\,\frac{128}{3}-20+d\\ =&\,\frac{68}{3}+d. \end{align*}

Hence

\begin{align*} d=-12. \end{align*}

Therefore

\begin{align*} f(x)=\frac{16}{3}x^{3/2}-\frac32x^2+x-12. \end{align*}