题目
Problem
(i) f ( x ) = x − 4 − cos ( 5 x ) x > 0 f(x) = x - 4 - \cos(5\sqrt{x}) \quad x > 0 f ( x ) = x − 4 − cos ( 5 x ) x > 0
(a) Show that the equation f ( x ) = 0 f(x) = 0 f ( x ) = 0 α \alpha α [ 2.5 , 3.5 ] [2.5, 3.5] [ 2.5 , 3.5 ]
(2)
(b) Use linear interpolation once on the interval [ 2.5 , 3.5 ] [2.5, 3.5] [ 2.5 , 3.5 ] α \alpha α 2 2 2
(2)
(ii) g ( x ) = 1 10 x 2 − 1 2 x 2 + x − 11 x > 0 g(x) = \dfrac{1}{10}x^2 - \dfrac{1}{2x^2} + x - 11 \quad x > 0 g ( x ) = 10 1 x 2 − 2 x 2 1 + x − 11 x > 0
(a) Determine g ′ ( x ) g'(x) g ′ ( x )
(2)
The equation g ( x ) = 0 g(x) = 0 g ( x ) = 0 β \beta β [ 6 , 7 ] [6, 7] [ 6 , 7 ]
(b) Using x 0 = 6 x_0 = 6 x 0 = 6 β \beta β g ( x ) g(x) g ( x ) β \beta β 3 3 3
(2)
题目中文翻译
(i) f ( x ) = x − 4 − cos ( 5 x ) x > 0 f(x) = x - 4 - \cos(5\sqrt{x}) \quad x > 0 f ( x ) = x − 4 − cos ( 5 x ) x > 0
(a) 证明方程 f ( x ) = 0 f(x) = 0 f ( x ) = 0 [ 2.5 , 3.5 ] [2.5, 3.5] [ 2.5 , 3.5 ] α \alpha α
(b) 在区间 [ 2.5 , 3.5 ] [2.5, 3.5] [ 2.5 , 3.5 ] α \alpha α 2 2 2
(ii) g ( x ) = 1 10 x 2 − 1 2 x 2 + x − 11 x > 0 g(x) = \dfrac{1}{10}x^2 - \dfrac{1}{2x^2} + x - 11 \quad x > 0 g ( x ) = 10 1 x 2 − 2 x 2 1 + x − 11 x > 0
(a) 确定 g ′ ( x ) g'(x) g ′ ( x )
方程 g ( x ) = 0 g(x) = 0 g ( x ) = 0 [ 6 , 7 ] [6, 7] [ 6 , 7 ] β \beta β
(b) 使用 x 0 = 6 x_0 = 6 x 0 = 6 β \beta β g ( x ) g(x) g ( x ) β \beta β 3 3 3
解答
(i)(a)
解法一
思路
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分别计算区间两端的函数值。由于函数在正数范围内连续，而两端函数值异号，便可由连续函数的介值性质说明区间内存在根。
答题过程
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f ( 2.5 ) = 2.5 − 4 − cos ( 5 2.5 ) = − 1.44831 … < 0 , \begin{align*}
f(2.5)
=&\,2.5-4-\cos(5\sqrt{2.5})\\
=&\,-1.44831\ldots<0,
\end{align*} f ( 2.5 ) = = 2.5 − 4 − cos ( 5 2.5 ) − 1.44831 … < 0 , and
f ( 3.5 ) = 3.5 − 4 − cos ( 5 3.5 ) = 0.497506 … > 0. \begin{align*}
f(3.5)
=&\,3.5-4-\cos(5\sqrt{3.5})\\
=&\,0.497506\ldots>0.
\end{align*} f ( 3.5 ) = = 3.5 − 4 − cos ( 5 3.5 ) 0.497506 … > 0. Since f f f x > 0 x>0 x > 0 f ( 2.5 ) f ( 3.5 ) < 0 f(2.5)f(3.5)<0 f ( 2.5 ) f ( 3.5 ) < 0
α ∈ [ 2.5 , 3.5 ] . \boxed{\alpha\in[2.5,3.5]}. α ∈ [ 2.5 , 3.5 ] .
(i)(b)
解法一
思路
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在两个端点对应的图像点之间作割线，并求这条割线与 x x x
答题过程
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Using linear interpolation once,
α ≈ 2.5 + 0 − f ( 2.5 ) f ( 3.5 ) − f ( 2.5 ) ( 3.5 − 2.5 ) = 2.5 + 1.448310 … 0.497506 … + 1.448310 … = 3.244320 … \begin{align*}
\alpha
\approx&\,2.5+
\frac{0-f(2.5)}{f(3.5)-f(2.5)}(3.5-2.5)\\
=&\,2.5+
\frac{1.448310\ldots}
{0.497506\ldots+1.448310\ldots}\\
=&\,3.244320\ldots
\end{align*} α ≈ = = 2.5 + f ( 3.5 ) − f ( 2.5 ) 0 − f ( 2.5 ) ( 3.5 − 2.5 ) 2.5 + 0.497506 … + 1.448310 … 1.448310 … 3.244320 … Therefore, to 2 decimal places,
α ≈ 3.24 . \boxed{\alpha\approx3.24}. α ≈ 3.24 .
(ii)(a)
解法一
思路
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先把分式项写成负指数幂，再逐项求导；常数项的导数为零。
答题过程
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Writing
g ( x ) = 1 10 x 2 − 1 2 x − 2 + x − 11 , g(x)=\frac1{10}x^2-\frac12x^{-2}+x-11, g ( x ) = 10 1 x 2 − 2 1 x − 2 + x − 11 , we obtain
g ′ ( x ) = 1 5 x + ( − 1 2 ) ( − 2 ) x − 3 + 1 = 1 5 x + 1 x 3 + 1 . \begin{align*}
g'(x)
=&\,\frac15x+(-\tfrac12)(-2)x^{-3}+1\\
=&\,\boxed{\frac15x+\frac1{x^3}+1}.
\end{align*} g ′ ( x ) = = 5 1 x + ( − 2 1 ) ( − 2 ) x − 3 + 1 5 1 x + x 3 1 + 1 .
(ii)(b)
解法一
思路
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使用 Newton–Raphson 迭代式 x n + 1 = x n − g ( x n ) / g ′ ( x n ) x_{n+1}=x_n-g(x_n)/g'(x_n) x n + 1 = x n − g ( x n ) / g ′ ( x n ) x 0 = 6 x_0=6 x 0 = 6
答题过程
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The Newton–Raphson formula is
x n + 1 = x n − g ( x n ) g ′ ( x n ) . x_{n+1}=x_n-\frac{g(x_n)}{g'(x_n)}. x n + 1 = x n − g ′ ( x n ) g ( x n ) . At x 0 = 6 x_0=6 x 0 = 6
g ( 6 ) = 1 10 ( 6 ) 2 − 1 2 ( 6 ) 2 + 6 − 11 = − 1.413888 … , \begin{align*}
g(6)
=&\,\frac1{10}(6)^2-\frac1{2(6)^2}+6-11\\
=&\,-1.413888\ldots,
\end{align*} g ( 6 ) = = 10 1 ( 6 ) 2 − 2 ( 6 ) 2 1 + 6 − 11 − 1.413888 … , and
g ′ ( 6 ) = 1 5 ( 6 ) + 1 6 3 + 1 = 2.204629 … . \begin{align*}
g'(6)
=&\,\frac15(6)+\frac1{6^3}+1\\
=&\,2.204629\ldots.
\end{align*} g ′ ( 6 ) = = 5 1 ( 6 ) + 6 3 1 + 1 2.204629 … . Therefore,
x 1 = 6 − − 1.413888 … 2.204629 … = 6.641327 … . \begin{align*}
x_1
=&\,6-\frac{-1.413888\ldots}{2.204629\ldots}\\
=&\,6.641327\ldots.
\end{align*} x 1 = = 6 − 2.204629 … − 1.413888 … 6.641327 … . Hence the second approximation, to 3 decimal places, is
x 1 = 6.641 . \boxed{x_1=6.641}. x 1 = 6.641 . 