Question
Problem
f ( x ) = 30 + 7 x − x 5 f(x) = 30 + \dfrac{7}{\sqrt{x}}-x^5 f ( x ) = 30 + x 7 − x 5 , x > 0 x > 0 x > 0
The only real root, α \alpha α , of the equation f ( x ) = 0 f(x) = 0 f ( x ) = 0 lies in the interval [ 2 , 2.1 ] [2, 2.1] [ 2 , 2.1 ] .
(a) Starting with the interval [ 2 , 2.1 ] [2, 2.1] [ 2 , 2.1 ] , use interval bisection twice to find an interval of width 0.025 that contains α \alpha α .
(4)
(b) Find f ′ ( x ) f'(x) f ′ ( x ) .
(2)
(c) Taking 2 as a first approximation to α \alpha α , apply the Newton-Raphson process once to f ( x ) f(x) f ( x ) to find a second approximation to α \alpha α , giving your answer to 2 decimal places.
(2)
中文翻译
f ( x ) = 30 + 7 x − x 5 f(x) = 30 + \dfrac{7}{\sqrt{x}}-x^5 f ( x ) = 30 + x 7 − x 5 ,x > 0 x > 0 x > 0
方程 f ( x ) = 0 f(x) = 0 f ( x ) = 0 的唯一实根 α \alpha α 在区间 [ 2 , 2.1 ] [2, 2.1] [ 2 , 2.1 ] 内。
(a) 从区间 [ 2 , 2.1 ] [2, 2.1] [ 2 , 2.1 ] 开始,使用区间二分法两次,找到宽度为 0.025 且包含 α \alpha α 的区间。
(4)
(b) 求 f ′ ( x ) f'(x) f ′ ( x ) 。
(2)
(c) 取 2 作为 α \alpha α 的初始近似值,对 f ( x ) f(x) f ( x ) 应用一次牛顿-拉夫森过程求 α \alpha α 的第二个近似值,答案精确到 2 位小数。
(2)
解答
(a)
解法一
思路
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每次二分都计算当前区间中点的函数值,并与一个端点的函数值比较符号。连续二分两次后,区间宽度由 0.1 0.1 0.1 变为 0.05 0.05 0.05 ,再变为 0.025 0.025 0.025 。
答题过程
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At the endpoints of the initial interval,
f ( 2 ) = 2.9497 … > 0 , f ( 2.1 ) = − 6.0105 … < 0. \begin{align*}
f(2)=&\,2.9497\ldots>0,\\
f(2.1)=&\,-6.0105\ldots<0.
\end{align*} f ( 2 ) = f ( 2.1 ) = 2.9497 … > 0 , − 6.0105 … < 0.
The first midpoint is 2.05 2.05 2.05 , and
f ( 2.05 ) = − 1.3160 … < 0. f(2.05)=-1.3160\ldots<0. f ( 2.05 ) = − 1.3160 … < 0.
Since f ( 2 ) > 0 f(2)>0 f ( 2 ) > 0 and f ( 2.05 ) < 0 f(2.05)<0 f ( 2.05 ) < 0 , the root lies in
[ 2 , 2.05 ] . [2,2.05]. [ 2 , 2.05 ] .
The midpoint of this new interval is 2.025 2.025 2.025 , and
f ( 2.025 ) = 0.8684 … > 0. f(2.025)=0.8684\ldots>0. f ( 2.025 ) = 0.8684 … > 0.
Since f ( 2.025 ) > 0 f(2.025)>0 f ( 2.025 ) > 0 and f ( 2.05 ) < 0 f(2.05)<0 f ( 2.05 ) < 0 ,
2.025 < α < 2.05 . \boxed{2.025<\alpha<2.05}. 2.025 < α < 2.05 .
This interval has width 2.05 − 2.025 = 0.025 2.05-2.025=0.025 2.05 − 2.025 = 0.025 .
(b)
解法一
思路
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先把根式分母写成幂 x − 1 / 2 x^{-1/2} x − 1/2 ,再逐项求导,注意负指数求导后的系数与指数符号。
答题过程
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Writing
f ( x ) = 30 + 7 x − 1 / 2 − x 5 , f(x)=30+7x^{-1/2}-x^5, f ( x ) = 30 + 7 x − 1/2 − x 5 ,
we obtain
f ′ ( x ) = − 7 2 x − 3 / 2 − 5 x 4 . \boxed{
f'(x)=-\frac{7}{2}x^{-3/2}-5x^4
}. f ′ ( x ) = − 2 7 x − 3/2 − 5 x 4 .
(c)
解法一
思路
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使用牛顿-拉夫森公式 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 ) ,只迭代一次。题目指定 x 0 = 2 x_0=2 x 0 = 2 ,所以代入 (a)、(b) 已算出的函数值和导数值即可。
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 ) .
Using x 0 = 2 x_0=2 x 0 = 2 ,
f ( 2 ) = 2.9497 … f(2)=2.9497\ldots f ( 2 ) = 2.9497 …
and
f ′ ( 2 ) = − 81.2374 … . f'(2)=-81.2374\ldots. f ′ ( 2 ) = − 81.2374 … .
Therefore,
x 1 = 2 − f ( 2 ) f ′ ( 2 ) = 2 − 2.9497 … − 81.2374 … = 2.0363 … \begin{align*}
x_1
=&\,2-\frac{f(2)}{f'(2)}\\
=&\,2-\frac{2.9497\ldots}{-81.2374\ldots}\\
=&\,2.0363\ldots
\end{align*} x 1 = = = 2 − f ′ ( 2 ) f ( 2 ) 2 − − 81.2374 … 2.9497 … 2.0363 …
Hence the second approximation, to two decimal places, is
x 1 = 2.04 . \boxed{x_1=2.04}. x 1 = 2.04 .