题目
Problem
2. f ( x ) = 10 − 2 x − 1 2 x − 1 x 3 x > 0 f(x) = 10 - 2x - \dfrac{1}{2\sqrt{x}} - \dfrac{1}{x^3} \quad x > 0 f ( x ) = 10 − 2 x − 2 x 1 − x 3 1 x > 0
(a) Show that the equation f ( x ) = 0 f(x) = 0 f ( x ) = 0 α \alpha α [ 0.4 , 0.5 ] [0.4, 0.5] [ 0.4 , 0.5 ]
(2)
(b) Determine f ′ ( x ) f'(x) f ′ ( x )
(3)
(c) Using x 0 = 0.5 x_0 = 0.5 x 0 = 0.5 α \alpha α f ( x ) f(x) f ( x ) α \alpha α
(2)
The equation f ( x ) = 0 f(x) = 0 f ( x ) = 0 β \beta β [ 4.8 , 4.9 ] [4.8, 4.9] [ 4.8 , 4.9 ]
(d) Use linear interpolation once on the interval [ 4.8 , 4.9 ] [4.8, 4.9] [ 4.8 , 4.9 ] β \beta β
(2)
题目中文翻译
f ( x ) = 10 − 2 x − 1 2 x − 1 x 3 x > 0 f(x) = 10 - 2x - \dfrac{1}{2\sqrt{x}} - \dfrac{1}{x^3} \quad x > 0 f ( x ) = 10 − 2 x − 2 x 1 − x 3 1 x > 0
(a) 证明方程 f ( x ) = 0 f(x) = 0 f ( x ) = 0 [ 0.4 , 0.5 ] [0.4, 0.5] [ 0.4 , 0.5 ] α \alpha α
(b) 求 f ′ ( x ) f'(x) f ′ ( x )
(c) 取 x 0 = 0.5 x_0 = 0.5 x 0 = 0.5 α \alpha α f ( x ) f(x) f ( x ) α \alpha α
方程 f ( x ) = 0 f(x) = 0 f ( x ) = 0 [ 4.8 , 4.9 ] [4.8, 4.9] [ 4.8 , 4.9 ] β \beta β
(d) 在区间 [ 4.8 , 4.9 ] [4.8, 4.9] [ 4.8 , 4.9 ] β \beta β
解答
(a)
解法一
思路
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分别计算区间两端的函数值。两者异号，并且 f f f
答题过程
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f ( 0.4 ) = − 7.2155 … < 0 , f ( 0.5 ) = 0.2928 … > 0. \begin{align*}
f(0.4)=&\,-7.2155\ldots<0,\\
f(0.5)=&\,0.2928\ldots>0.
\end{align*} f ( 0.4 ) = f ( 0.5 ) = − 7.2155 … < 0 , 0.2928 … > 0. Since f f f [ 0.4 , 0.5 ] [0.4,0.5] [ 0.4 , 0.5 ]
α ∈ ( 0.4 , 0.5 ) . \boxed{\alpha\in(0.4,0.5)}. α ∈ ( 0.4 , 0.5 ) .
(b)
解法一
思路
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先把根式与分式写成幂：f ( x ) = 10 − 2 x − 1 2 x − 1 / 2 − x − 3 f(x)=10-2x-\frac12x^{-1/2}-x^{-3} f ( x ) = 10 − 2 x − 2 1 x − 1/2 − x − 3
答题过程
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Writing f f f
f ( x ) = 10 − 2 x − 1 2 x − 1 / 2 − x − 3 . f(x)=10-2x-\frac12x^{-1/2}-x^{-3}. f ( x ) = 10 − 2 x − 2 1 x − 1/2 − x − 3 . Therefore,
f ′ ( x ) = − 2 + 1 4 x − 3 / 2 + 3 x − 4 = − 2 + 1 4 x 3 / 2 + 3 x 4 . \begin{align*}
f'(x)
=&\,-2+\frac14x^{-3/2}+3x^{-4}\\
=&\,\boxed{-2+\frac{1}{4x^{3/2}}+\frac3{x^4}}.
\end{align*} f ′ ( x ) = = − 2 + 4 1 x − 3/2 + 3 x − 4 − 2 + 4 x 3/2 1 + x 4 3 .
(c)
解法一
思路
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使用 Newton-Raphson 公式 x n + 1 = x n − f ( x n ) / f ′ ( x n ) x_{n+1}=x_n-f(x_n)/f'(x_n) x n + 1 = x n − f ( x n ) / f ′ ( x n ) x 0 = 0.5 x_0=0.5 x 0 = 0.5
答题过程
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The Newton-Raphson formula is
x n + 1 = x n − f ( x n ) f ′ ( x n ) . x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}. x n + 1 = x n − f ′ ( x n ) f ( x n ) . At x 0 = 0.5 x_0=0.5 x 0 = 0.5
f ( 0.5 ) = 0.29289321 … , f ′ ( 0.5 ) = 46.70710678 … . \begin{align*}
f(0.5)=&\,0.29289321\ldots,\\
f'(0.5)=&\,46.70710678\ldots.
\end{align*} f ( 0.5 ) = f ′ ( 0.5 ) = 0.29289321 … , 46.70710678 … . Hence
x 1 = 0.5 − 0.29289321 … 46.70710678 … = 0.49372915 … = 0.494 (to 3 d.p.). \begin{align*}
x_1
=&\,0.5-\frac{0.29289321\ldots}
{46.70710678\ldots}\\
=&\,0.49372915\ldots\\
=&\,\boxed{0.494}\quad\text{(to 3 d.p.).}
\end{align*} x 1 = = = 0.5 − 46.70710678 … 0.29289321 … 0.49372915 … 0.494 (to 3 d.p.).
(d)
解法一
思路
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先计算端点函数值，再把两点 ( 4.8 , f ( 4.8 ) ) (4.8,f(4.8)) ( 4.8 , f ( 4.8 )) ( 4.9 , f ( 4.9 ) ) (4.9,f(4.9)) ( 4.9 , f ( 4.9 )) 0.1 0.1 0.1
答题过程
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f ( 4.8 ) = 0.162740 … , f ( 4.9 ) = − 0.0343768 … . \begin{align*}
f(4.8)=&\,0.162740\ldots,\\
f(4.9)=&\,-0.0343768\ldots.
\end{align*} f ( 4.8 ) = f ( 4.9 ) = 0.162740 … , − 0.0343768 … . Using linear interpolation,
β − 4.8 4.9 − 4.8 = 0.162740 … 0.162740 … + 0.0343768 … . \frac{\beta-4.8}{4.9-4.8}
=\frac{0.162740\ldots}
{0.162740\ldots+0.0343768\ldots}. 4.9 − 4.8 β − 4.8 = 0.162740 … + 0.0343768 … 0.162740 … . Therefore,
β = 4.8 + 0.1 ( 0.162740 … 0.197116 … ) = 4.882560 … = 4.883 (to 3 d.p.). \begin{align*}
\beta
=&\,4.8+0.1
\left(\frac{0.162740\ldots}{0.197116\ldots}\right)\\
=&\,4.882560\ldots\\
=&\,\boxed{4.883}\quad\text{(to 3 d.p.).}
\end{align*} β = = = 4.8 + 0.1 ( 0.197116 … 0.162740 … ) 4.882560 … 4.883 (to 3 d.p.). 