题目
Problem
2.
f ( x ) = 10 − 2 x − 1 2 x − 1 x 3 \begin{align*}
&\,f(x)\\
=&\,10-2x\\
&\,\hspace{2pt}-\frac{1}{2\sqrt{x}}-\frac{1}{x^3}
\end{align*} = f ( x ) 10 − 2 x − 2 x 1 − x 3 1 x > 0 x > 0 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 \begin{align*}
&\,f(x)\\
=&\,10-2x\\
&\,\hspace{2pt}-\frac{1}{2\sqrt{x}}-\frac{1}{x^3}
\end{align*} = f ( x ) 10 − 2 x − 2 x 1 − x 3 1 x > 0 x > 0 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
(2)
(b) 求 f ′ ( x ) f'(x) f ′ ( x )
(3)
(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
(2)
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
(2)
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 ) = 10 − 2 ( 0.4 ) − 1 2 0.4 − 1 0.4 3 = − 7.215569 … < 0. \begin{align*}
&\,f(0.4)\\
=&\,10-2(0.4)\\
&\,\hspace{2pt}-\frac1{2\sqrt{0.4}}-\frac1{0.4^3}\\
=&\,-7.215569\ldots<0.
\end{align*} = = f ( 0.4 ) 10 − 2 ( 0.4 ) − 2 0.4 1 − 0. 4 3 1 − 7.215569 … < 0. f ( 0.5 ) = 10 − 2 ( 0.5 ) − 1 2 0.5 − 1 0.5 3 = 0.292893 … > 0. \begin{align*}
&\,f(0.5)\\
=&\,10-2(0.5)\\
&\,\hspace{2pt}-\frac1{2\sqrt{0.5}}-\frac1{0.5^3}\\
=&\,0.292893\ldots>0.
\end{align*} = = f ( 0.5 ) 10 − 2 ( 0.5 ) − 2 0.5 1 − 0. 5 3 1 0.292893 … > 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，再逐项求导。
答题过程
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Writing f f f
f ( x ) = 10 − 2 x − 1 2 x − 1 / 2 − x − 3 . \begin{align*}
f(x)=&\,10-2x\\
&\,-\frac12x^{-1/2}-x^{-3}.
\end{align*} f ( x ) = 10 − 2 x − 2 1 x − 1/2 − x − 3 . Therefore,
f ′ ( x ) = − 2 − 1 2 ( − 1 2 ) x − 3 / 2 − ( − 3 ) x − 4 = − 2 + 1 4 x − 3 / 2 + 3 x − 4 . \begin{align*}
&\,f'(x)\\
=&\,-2\\
&\,\hspace{2pt}-\frac12\big(-\frac12\big)x^{-3/2}\\
&\,\hspace{4pt}-(-3)x^{-4}\\
=&\,-2+\frac14x^{-3/2}+3x^{-4}.
\end{align*} = = f ′ ( x ) − 2 − 2 1 ( − 2 1 ) x − 3/2 − ( − 3 ) x − 4 − 2 + 4 1 x − 3/2 + 3 x − 4 . Equivalently,
f ′ ( x ) = − 2 + 1 4 x 3 / 2 + 3 x 4 . \boxed{\begin{aligned}
f'(x)=&\,-2+\frac{1}{4x^{3/2}}\\
&\,+\frac3{x^4}
\end{aligned}}. f ′ ( x ) = − 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，代入 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 . \begin{align*}
&\,x_1\\
=&\,0.5-\frac{0.29289321\ldots}
{46.70710678\ldots}\\
=&\,0.49372915\ldots\\
=&\,\boxed{0.494}.
\end{align*} = = = x 1 0.5 − 46.70710678 … 0.29289321 … 0.49372915 … 0.494 . This is the second approximation to α \alpha α
(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 ) = 10 − 2 ( 4.8 ) − 1 2 4.8 − 1 4.8 3 = 0.162740022 … . \begin{align*}
&\,f(4.8)\\
=&\,10-2(4.8)\\
&\,\hspace{2pt}-\frac1{2\sqrt{4.8}}-\frac1{4.8^3}\\
=&\,0.162740022\ldots.
\end{align*} = = f ( 4.8 ) 10 − 2 ( 4.8 ) − 2 4.8 1 − 4. 8 3 1 0.162740022 … . f ( 4.9 ) = 10 − 2 ( 4.9 ) − 1 2 4.9 − 1 4.9 3 = − 0.034376835 … . \begin{align*}
&\,f(4.9)\\
=&\,10-2(4.9)\\
&\,\hspace{2pt}-\frac1{2\sqrt{4.9}}-\frac1{4.9^3}\\
=&\,-0.034376835\ldots.
\end{align*} = = f ( 4.9 ) 10 − 2 ( 4.9 ) − 2 4.9 1 − 4. 9 3 1 − 0.034376835 … . Let D D D
D = f ( 4.8 ) − f ( 4.9 ) = 0.162740022 … + 0.034376835 … = 0.197116857 … . \begin{align*}
D=&\,f(4.8)-f(4.9)\\
=&\,0.162740022\ldots\\
&\,\hspace{2pt}+0.034376835\ldots\\
=&\,0.197116857\ldots.
\end{align*} D = = = f ( 4.8 ) − f ( 4.9 ) 0.162740022 … + 0.034376835 … 0.197116857 … . Using linear interpolation,
β − 4.8 4.9 − 4.8 = 0.162740022 … D . \begin{align*}
&\,\frac{\beta-4.8}{4.9-4.8}\\
=&\,\frac{0.162740022\ldots}{D}.
\end{align*} = 4.9 − 4.8 β − 4.8 D 0.162740022 … . Therefore,
β = 4.8 + 0.1 × ( 0.162740022 … D ) = 4.882560 … = 4.883 . \begin{align*}
&\,\beta\\
=&\,4.8+0.1\\
&\,\hspace{2pt}\times\bigg(\frac{0.162740022\ldots}{D}\bigg)\\
=&\,4.882560\ldots\\
=&\,\boxed{4.883}.
\end{align*} = = = β 4.8 + 0.1 × ( D 0.162740022 … ) 4.882560 … 4.883 . The answer is given to 3 decimal places; unrounded endpoint values were used in the calculation.
解法二
思路
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把区间两端的函数值看作两个点的纵坐标，先求连接它们的直线斜率，再令直线的纵坐标为零，得到横截距。
答题过程
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The line joins the points
( 4.8 , f ( 4.8 ) ) , ( 4.9 , f ( 4.9 ) ) , \begin{align*}
&\,\big(4.8,f(4.8)\big),\\
&\,\big(4.9,f(4.9)\big),
\end{align*} ( 4.8 , f ( 4.8 ) ) , ( 4.9 , f ( 4.9 ) ) , where, as calculated above,
f ( 4.8 ) = 0.162740022 … , f ( 4.9 ) = − 0.034376835 … . \begin{align*}
&\,f(4.8)\\
=&\,0.162740022\ldots,\\
&\,f(4.9)\\
=&\,-0.034376835\ldots.
\end{align*} = = f ( 4.8 ) 0.162740022 … , f ( 4.9 ) − 0.034376835 … . Its gradient is
m = f ( 4.9 ) − f ( 4.8 ) 4.9 − 4.8 = − 0.197116857 … 0.1 = − 1.971168578 … . \begin{align*}
m=&\,\frac{f(4.9)-f(4.8)}{4.9-4.8}\\
=&\,\frac{-0.197116857\ldots}{0.1}\\
=&\,-1.971168578\ldots.
\end{align*} m = = = 4.9 − 4.8 f ( 4.9 ) − f ( 4.8 ) 0.1 − 0.197116857 … − 1.971168578 … . Using the point at x = 4.8 x=4.8 x = 4.8
y − f ( 4.8 ) = m ( x − 4.8 ) . y-f(4.8)=m(x-4.8). y − f ( 4.8 ) = m ( x − 4.8 ) . At the horizontal intercept, y = 0 y=0 y = 0 x = β x=\beta x = β
− f ( 4.8 ) = m ( β − 4.8 ) , β = 4.8 − f ( 4.8 ) m = 4.8 − 0.162740 … − 1.971168 … = 4.882560174 … = 4.883 . \begin{align*}
&\,-f(4.8)=m(\beta-4.8),\\
&\,\beta\\
=&\,4.8-\frac{f(4.8)}m\\
=&\,4.8-\frac{0.162740\ldots}{-1.971168\ldots}\\
=&\,4.882560174\ldots\\
=&\,\boxed{4.883}.
\end{align*} = = = = − f ( 4.8 ) = m ( β − 4.8 ) , β 4.8 − m f ( 4.8 ) 4.8 − − 1.971168 … 0.162740 … 4.882560174 … 4.883 . The answer is given to 3 decimal places; unrounded values were used throughout.
