题目
Problem
f ( x ) = 3 x − 1 x − 5 f(x)=3\sqrt{x}-\frac1x-5 f ( x ) = 3 x − x 1 − 5
where x > 0 x>0 x > 0 .
The equation f ( x ) = 0 f(x) = 0 f ( x ) = 0 has a root α \alpha α in the interval [ 3 , 3.8 ] [3, 3.8] [ 3 , 3.8 ]
(a) Starting with the end points of this interval, use interval bisection to determine α \alpha α to one decimal place.
(4)
(b) (i) Determine f ′ ( x ) f'(x) f ′ ( x ) .
(2)
(ii) Using your answer to part (a) as a first approximation to α \alpha α , apply the Newton-Raphson process once to calculate a further approximation for α \alpha α , giving your answer to 4 decimal places.
(2)
题目中文翻译
f ( x ) = 3 x − 1 x − 5 f(x)=3\sqrt{x}-\frac1x-5 f ( x ) = 3 x − x 1 − 5
其中 x > 0 x>0 x > 0 。
方程 f ( x ) = 0 f(x) = 0 f ( x ) = 0 在区间 [ 3 , 3.8 ] [3, 3.8] [ 3 , 3.8 ] 内有根 α \alpha α
(a) 从该区间的端点开始,使用区间二分法将 α \alpha α 确定到 1 位小数。
(4)
(b) (i) 求 f ′ ( x ) f'(x) f ′ ( x ) 。
(2)
(ii) 使用 (a) 的答案作为 α \alpha α 的初始近似值,应用一次牛顿-拉夫森过程计算 α \alpha α 的进一步近似值,答案精确到 4 位小数。
(2)
解答
(a)
解法一
思路
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每次计算当前区间中点的函数值,并利用端点函数值异号缩小含根区间。为了确定到 1 位小数,最后还要检查舍入分界点 3.15 3.15 3.15 ,确认根位于其左侧。
答题过程
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At the initial endpoints,
f ( 3 ) = − 0.1371809 … f(3)=-0.1371809\ldots f ( 3 ) = − 0.1371809 …
and
f ( 3.8 ) = 0.5849187 … . f(3.8)=0.5849187\ldots. f ( 3.8 ) = 0.5849187 … .
The first midpoint is 3.4 3.4 3.4 :
f ( 3.4 ) = 0.2376090 … > 0. \begin{align*}
&\,f(3.4)\\
=&\,0.2376090\ldots>0.
\end{align*} = f ( 3.4 ) 0.2376090 … > 0.
The new interval is [ 3 , 3.4 ] [3,3.4] [ 3 , 3.4 ] . Its midpoint is 3.2 3.2 3.2 :
f ( 3.2 ) = 0.0540631 … > 0. \begin{align*}
&\,f(3.2)\\
=&\,0.0540631\ldots>0.
\end{align*} = f ( 3.2 ) 0.0540631 … > 0.
The new interval is [ 3 , 3.2 ] [3,3.2] [ 3 , 3.2 ] . Its midpoint is 3.1 3.1 3.1 :
f ( 3.1 ) = − 0.0405355 … < 0. \begin{align*}
&\,f(3.1)\\
=&\,-0.0405355\ldots<0.
\end{align*} = f ( 3.1 ) − 0.0405355 … < 0.
The new interval is [ 3.1 , 3.2 ] [3.1,3.2] [ 3.1 , 3.2 ] . Its midpoint is 3.15 3.15 3.15 :
f ( 3.15 ) = 0.0070114 … > 0. \begin{align*}
&\,f(3.15)\\
=&\,0.0070114\ldots>0.
\end{align*} = f ( 3.15 ) 0.0070114 … > 0.
Thus the final interval is [ 3.1 , 3.15 ] [3.1,3.15] [ 3.1 , 3.15 ] .
Thus the root lies below the rounding boundary 3.15 3.15 3.15 , so to one decimal place,
α ≈ 3.1 . \boxed{\alpha\approx3.1}. α ≈ 3.1 .
(b)(i)
解法一
思路
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把根式和倒数分别写成 x 1 / 2 x^{1/2} x 1/2 与 x − 1 x^{-1} x − 1 ,再逐项使用幂函数求导法则。注意 − x − 1 -x^{-1} − x − 1 求导后变成正的 x − 2 x^{-2} x − 2 。
答题过程
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Writing
f ( x ) = 3 x 1 / 2 − x − 1 − 5 , f(x)=3x^{1/2}-x^{-1}-5, f ( x ) = 3 x 1/2 − x − 1 − 5 ,
we obtain
f ′ ( x ) = 3 2 x − 1 / 2 + x − 2 = 3 2 x + 1 x 2 . \begin{align*}
f'(x)
=&\,\frac32x^{-1/2}+x^{-2}\\
=&\,\boxed{\frac{3}{2\sqrt{x}}+\frac1{x^2}}.
\end{align*} f ′ ( x ) = = 2 3 x − 1/2 + x − 2 2 x 3 + x 2 1 .
(b)(ii)
解法一
思路
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使用 (a) 的近似值 x 0 = 3.1 x_0=3.1 x 0 = 3.1 ,代入 Newton–Raphson 公式 x 1 = x 0 − f ( x 0 ) f ′ ( x 0 ) x_1=x_0-\frac{f(x_0)}{f'(x_0)} x 1 = x 0 − f ′ ( x 0 ) f ( x 0 ) ,只迭代一次并在最后保留 4 位小数。
答题过程
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Using x 0 = 3.1 x_0=3.1 x 0 = 3.1 from part (a),
f ( 3.1 ) = 3 3.1 − 1 3.1 − 5 = − 0.0405355 … , \begin{align*}
&\,f(3.1)\\
=&\,3\sqrt{3.1}-\frac1{3.1}-5\\
=&\,-0.0405355\ldots,
\end{align*} = = f ( 3.1 ) 3 3.1 − 3.1 1 − 5 − 0.0405355 … ,
and
f ′ ( 3.1 ) = 3 2 3.1 + 1 ( 3.1 ) 2 . \begin{align*}
f'(3.1)=&\,\frac3{2\sqrt{3.1}}+\frac1{(3.1)^2}.
\end{align*} f ′ ( 3.1 ) = 2 3.1 3 + ( 3.1 ) 2 1 .
Hence,
x 1 = x 0 − f ( x 0 ) f ′ ( x 0 ) = 3.1 − f ( 3.1 ) f ′ ( 3.1 ) = 3.1424011 … \begin{align*}
&\,x_1\\
=&\,x_0-\frac{f(x_0)}{f'(x_0)}\\
=&\,3.1-\frac{f(3.1)}{f'(3.1)}\\
=&\,3.1424011\ldots
\end{align*} = = = x 1 x 0 − f ′ ( x 0 ) f ( x 0 ) 3.1 − f ′ ( 3.1 ) f ( 3.1 ) 3.1424011 …
Therefore, the further approximation is
α ≈ 3.1424 \boxed{\alpha\approx3.1424} α ≈ 3.1424
to 4 decimal places.