题目
Problem
f ( x ) = x 3 − 5 x − 4 x + 7 x ≥ 0 f(x) = x^3 - 5\sqrt{x} - 4x + 7 \quad x \geq 0 f ( x ) = x 3 − 5 x − 4 x + 7 x ≥ 0
The equation f ( x ) = 0 f(x) = 0 f ( x ) = 0 α \alpha α [ 0.25 , 1 ] [0.25, 1] [ 0.25 , 1 ]
(a) Use linear interpolation once on the interval [ 0.25 , 1 ] [0.25, 1] [ 0.25 , 1 ] α \alpha α
(3)
The equation f ( x ) = 0 f(x) = 0 f ( x ) = 0 β \beta β [ 1.5 , 2.5 ] [1.5, 2.5] [ 1.5 , 2.5 ]
(b) Determine f ′ ( x ) f'(x) f ′ ( x )
(2)
(c) Hence, using x 0 = 1.75 x_0 = 1.75 x 0 = 1.75 β \beta β f ( x ) f(x) f ( x ) β \beta β
(2)
题目中文翻译
已知函数
f ( x ) = x 3 − 5 x − 4 x + 7 x ≥ 0 f(x) = x^3 - 5\sqrt{x} - 4x + 7 \quad x \geq 0 f ( x ) = x 3 − 5 x − 4 x + 7 x ≥ 0
方程 f ( x ) = 0 f(x) = 0 f ( x ) = 0 [ 0.25 , 1 ] [0.25, 1] [ 0.25 , 1 ] α \alpha α
(a) 在区间 [ 0.25 , 1 ] [0.25, 1] [ 0.25 , 1 ] α \alpha α
方程 f ( x ) = 0 f(x) = 0 f ( x ) = 0 [ 1.5 , 2.5 ] [1.5, 2.5] [ 1.5 , 2.5 ] β \beta β
(b) 求出 f ′ ( x ) f'(x) f ′ ( x )
(c) 以 x 0 = 1.75 x_0 = 1.75 x 0 = 1.75 β \beta β f ( x ) f(x) f ( x ) β \beta β
解答
(a)
解法一
思路
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先计算区间两端的函数值，再把两端点之间的曲线近似为直线。使用线性插值公式求这条直线与 x x x
答题过程
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At the endpoints,
f ( 0.25 ) = 3.515625 f(0.25)=3.515625 f ( 0.25 ) = 3.515625 and
f ( 1 ) = − 1. f(1)=-1. f ( 1 ) = − 1. Using linear interpolation,
α = 0.25 f ( 1 ) − 1 f ( 0.25 ) f ( 1 ) − f ( 0.25 ) = 0.25 ( − 1 ) − 3.515625 − 1 − 3.515625 = 0.8339100346 … . \begin{align*}
\alpha
=&\,\frac{0.25f(1)-1f(0.25)}
{f(1)-f(0.25)}\\
=&\,\frac{0.25(-1)-3.515625}
{-1-3.515625}\\
=&\,0.8339100346\ldots.
\end{align*} α = = = f ( 1 ) − f ( 0.25 ) 0.25 f ( 1 ) − 1 f ( 0.25 ) − 1 − 3.515625 0.25 ( − 1 ) − 3.515625 0.8339100346 … . Therefore, to 3 decimal places,
α ≈ 0.834 . \boxed{\alpha\approx0.834}. α ≈ 0.834 .
(b)
解法一
思路
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先把根号写成 x 1 / 2 x^{1/2} x 1/2
答题过程
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Writing x = x 1 / 2 \sqrt{x}=x^{1/2} x = x 1/2
f ′ ( x ) = 3 x 2 − 5 2 x − 1 2 − 4. \begin{align*}
f'(x)
=&\,3x^2-\frac52x^{-\frac12}-4.
\end{align*} f ′ ( x ) = 3 x 2 − 2 5 x − 2 1 − 4. Hence,
f ′ ( x ) = 3 x 2 − 5 2 x − 4 . \boxed{f'(x)=3x^2-\frac{5}{2\sqrt{x}}-4}. f ′ ( x ) = 3 x 2 − 2 x 5 − 4 .
(c)
解法一
思路
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承接 (b) 的导函数，把 x 0 = 1.75 x_0=1.75 x 0 = 1.75
答题过程
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Using
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 ) , we have
f ( 1.75 ) = − 1.255003278 … f(1.75)=-1.255003278\ldots f ( 1.75 ) = − 1.255003278 … and
f ′ ( 1.75 ) = 3.297677635 … . f'(1.75)=3.297677635\ldots. f ′ ( 1.75 ) = 3.297677635 … . Therefore,
x 1 = 1.75 − − 1.255003278 … 3.297677635 … = 2.13057185 … . \begin{align*}
x_1
=&\,1.75-\frac{-1.255003278\ldots}
{3.297677635\ldots}\\
=&\,2.13057185\ldots.
\end{align*} x 1 = = 1.75 − 3.297677635 … − 1.255003278 … 2.13057185 … . Hence, to 3 decimal places,
β ≈ 2.131 . \boxed{\beta\approx2.131}. β ≈ 2.131 . 