题目
Problem
5.
f ( x ) = x 4 − 12 x 3 / 2 + 7 \begin{align*}
f(x)=&\,x^4\\
&\,\hspace{2pt}-12x^{3/2}+7
\end{align*} f ( x ) = x 4 − 12 x 3/2 + 7 x ≥ 0 x\geq0 x ≥ 0
(a) Show that the equation f ( x ) = 0 f(x) = 0 f ( x ) = 0 α \alpha α [ 2 , 3 ] [2, 3] [ 2 , 3 ]
(2)
(b) Taking 2.5 2.5 2.5 α \alpha α f ( x ) f(x) f ( x ) α \alpha α
(4)
(c) Show that your answer to (b) gives α \alpha α
(2)
题目中文翻译
5.
f ( x ) = x 4 − 12 x 3 / 2 + 7 \begin{align*}
f(x)=&\,x^4\\
&\,\hspace{2pt}-12x^{3/2}+7
\end{align*} f ( x ) = x 4 − 12 x 3/2 + 7 x ≥ 0 x\geq0 x ≥ 0
(a) 证明方程 f ( x ) = 0 f(x) = 0 f ( x ) = 0 [ 2 , 3 ] [2, 3] [ 2 , 3 ] α \alpha α
(2)
(b) 取 2.5 2.5 2.5 α \alpha α f ( x ) f(x) f ( x ) α \alpha α
(4)
(c) 证明 (b) 的答案给出 α \alpha α
(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。由于函数在 [ 2 , 3 ] [2,3] [ 2 , 3 ]
答题过程
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f ( 2 ) = 2 4 − 12 ( 2 ) 3 2 + 7 = 23 − 24 2 ≈ − 10.941 , \begin{align*}
f(2)
=&\,2^4-12(2)^{\frac{3}{2}}+7 \\
=&\,23-24\sqrt{2} \\
\approx&\,-10.941,
\end{align*} f ( 2 ) = = ≈ 2 4 − 12 ( 2 ) 2 3 + 7 23 − 24 2 − 10.941 , and
f ( 3 ) = 3 4 − 12 ( 3 ) 3 2 + 7 = 88 − 36 3 ≈ 25.646. \begin{align*}
f(3)
=&\,3^4-12(3)^{\frac{3}{2}}+7 \\
=&\,88-36\sqrt{3} \\
\approx&\,25.646.
\end{align*} f ( 3 ) = = ≈ 3 4 − 12 ( 3 ) 2 3 + 7 88 − 36 3 25.646. Thus f ( 2 ) < 0 < f ( 3 ) f(2)<0<f(3) f ( 2 ) < 0 < f ( 3 ) f f f [ 2 , 3 ] [2,3] [ 2 , 3 ]
α ∈ ( 2 , 3 ) . \boxed{\alpha\in(2,3)}. α ∈ ( 2 , 3 ) .
(b)
解法一
思路
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先求导，再把初值 x 0 = 2.5 x_0=2.5 x 0 = 2.5 x n + 1 = x n − f ( x n ) f ′ ( x n ) x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)} x n + 1 = x n − f ′ ( x n ) f ( x n )
答题过程
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Differentiating,
f ′ ( x ) = 4 x 3 − 18 x 1 2 . f'(x)=4x^3-18x^{\frac{1}{2}}. f ′ ( x ) = 4 x 3 − 18 x 2 1 . Using x 0 = 2.5 x_0=2.5 x 0 = 2.5
f ( 2.5 ) = ( 2.5 ) 4 − 12 ( 2.5 ) 3 2 + 7 = − 1.371664 … , \begin{align*}
&\,f(2.5)\\
=&\,(2.5)^4-12(2.5)^{\frac{3}{2}}+7 \\
=&\,-1.371664\ldots,
\end{align*} = = f ( 2.5 ) ( 2.5 ) 4 − 12 ( 2.5 ) 2 3 + 7 − 1.371664 … , and
f ′ ( 2.5 ) = 4 ( 2.5 ) 3 − 18 ( 2.5 ) 1 2 = 34.039501 … . \begin{align*}
&\,f'(2.5)\\
=&\,4(2.5)^3-18(2.5)^{\frac{1}{2}} \\
=&\,34.039501\ldots.
\end{align*} = = f ′ ( 2.5 ) 4 ( 2.5 ) 3 − 18 ( 2.5 ) 2 1 34.039501 … . Therefore,
x 1 = x 0 − f ( x 0 ) f ′ ( x 0 ) = 2.5 − − 1.371664 … 34.039501 … = 2.540296 … \begin{align*}
x_1
=&\,x_0-\frac{f(x_0)}{f'(x_0)} \\
=&\,2.5-\frac{-1.371664\ldots}{34.039501\ldots} \\
=&\,2.540296\ldots
\end{align*} x 1 = = = x 0 − f ′ ( x 0 ) f ( x 0 ) 2.5 − 34.039501 … − 1.371664 … 2.540296 … Therefore, to two decimal places,
α ≈ 2.54 . \boxed{\alpha\approx2.54}. α ≈ 2.54 .
(c)
解法一
思路
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一个数四舍五入到两位小数为 2.54 2.54 2.54 2.535 2.535 2.535 2.545 2.545 2.545
答题过程
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f ( 2.535 ) = − 0.137392 … < 0. \begin{align*}
f(2.535)=&\,-0.137392\ldots\\
<&\,0.
\end{align*} f ( 2.535 ) = < − 0.137392 … 0. and
f ( 2.545 ) = 0.231219 … > 0. \begin{align*}
f(2.545)=&\,0.231219\ldots\\
>&\,0.
\end{align*} f ( 2.545 ) = > 0.231219 … 0. Since f f f 2.535 2.535 2.535 2.545 2.545 2.545
2.535 < α < 2.545. 2.535<\alpha<2.545. 2.535 < α < 2.545. Every number in this interval rounds to 2.54 2.54 2.54
α = 2.54 \boxed{\alpha=2.54} α = 2.54 correct to two decimal places.
解法二
思路
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也可以再做一次 Newton–Raphson 迭代。若下一次近似仍为 2.538 … 2.538\ldots 2.538 … 2.54 2.54 2.54
答题过程
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Applying Newton–Raphson once more, starting from x 1 = 2.54 x_1=2.54 x 1 = 2.54
x 2 = 2.54 − f ( 2.54 ) f ′ ( 2.54 ) = 2.54 − 0.046016 … 36.860976 … = 2.538751 … \begin{align*}
x_2
=&\,2.54-\frac{f(2.54)}{f'(2.54)} \\
=&\,2.54-\frac{0.046016\ldots}{36.860976\ldots} \\
=&\,2.538751\ldots
\end{align*} x 2 = = = 2.54 − f ′ ( 2.54 ) f ( 2.54 ) 2.54 − 36.860976 … 0.046016 … 2.538751 … This also rounds to 2.54 2.54 2.54
α = 2.54 \boxed{\alpha=2.54} α = 2.54 correct to two decimal places.
