Question
Problem
f ( x ) = 3 x − 1 x − 5 f(x) = 3\sqrt{x} - \dfrac{1}{x} - 5 f ( x ) = 3 x − x 1 − 5 x > 0 x>0 x > 0
The equation f ( x ) = 0 f(x) = 0 f ( x ) = 0 α \alpha α [ 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
(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 α \alpha α
(2)
中文翻译
f ( x ) = 3 x − 1 x − 5 f(x) = 3\sqrt{x} - \dfrac{1}{x} - 5 f ( x ) = 3 x − x 1 − 5 x > 0 x>0 x > 0
方程 f ( x ) = 0 f(x) = 0 f ( x ) = 0 [ 3 , 3.8 ] [3, 3.8] [ 3 , 3.8 ] α \alpha α
(a) 从该区间的端点开始，使用区间二分法将 α \alpha α
(4)
(b) (i) 求 f ′ ( x ) f'(x) f ′ ( x )
(2)
(ii) 使用 (a) 的答案作为 α \alpha α α \alpha α
(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
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 successive bisection values are
x f ( x ) new interval containing α 3.4 0.2376090 … [ 3 , 3.4 ] 3.2 0.0540631 … [ 3 , 3.2 ] 3.1 − 0.0405355 … [ 3.1 , 3.2 ] 3.15 0.0070114 … [ 3.1 , 3.15 ] \begin{array}{c|c|c}
x & f(x) & \text{new interval containing }\alpha\\
\hline
3.4 & 0.2376090\ldots & [3,3.4]\\
3.2 & 0.0540631\ldots & [3,3.2]\\
3.1 & -0.0405355\ldots & [3.1,3.2]\\
3.15 & 0.0070114\ldots & [3.1,3.15]
\end{array} x 3.4 3.2 3.1 3.15 f ( x ) 0.2376090 … 0.0540631 … − 0.0405355 … 0.0070114 … new interval containing α [ 3 , 3.4 ] [ 3 , 3.2 ] [ 3.1 , 3.2 ] [ 3.1 , 3.15 ] Thus the root lies below the rounding boundary 3.15 3.15 3.15
α = 3.1 . \boxed{\alpha=3.1}. 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 x − 1 x^{-1} x − 1 − x − 1 -x^{-1} − x − 1 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 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 )
答题过程
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Using x 0 = 3.1 x_0=3.1 x 0 = 3.1
x 1 = x 0 − f ( x 0 ) f ′ ( x 0 ) = 3.1 − 3 3.1 − 1 3.1 − 5 3 2 3.1 + 1 ( 3.1 ) 2 = 3.1424011 … \begin{align*}
x_1
=&\,x_0-\frac{f(x_0)}{f'(x_0)}\\
=&\,3.1-
\frac{3\sqrt{3.1}-\frac1{3.1}-5}
{\frac{3}{2\sqrt{3.1}}+\frac1{(3.1)^2}}\\
=&\,3.1424011\ldots
\end{align*} x 1 = = = x 0 − f ′ ( x 0 ) f ( x 0 ) 3.1 − 2 3.1 3 + ( 3.1 ) 2 1 3 3.1 − 3.1 1 − 5 3.1424011 … Therefore, the further approximation is
α = 3.1424 \boxed{\alpha=3.1424} α = 3.1424 to 4 decimal places.
