Question
Problem
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
(i) f(x)=x2+5−85x
Given that the equation f(x)=0 has a single root, α, in the interval [0,1]
use interval bisection to determine an interval of width 0.25 that contains α.
(3)
(ii) g(x)=3sinx−3cosx
where x is in radians.
(a) Show that the equation g(x)=0 has a root, β, in the interval [4,5]
(2)
(b) Use linear interpolation once on the interval [4,5] to determine an approximation for β.
Give your answer to 4 significant figures.
(2)
中文翻译
本题必须展示所有运算步骤。
完全依赖计算器技术的解答不可接受。
(i) f(x)=x2+5−85x
已知方程 f(x)=0 在区间 [0,1] 内有唯一根 α
使用区间二分法确定宽度为 0.25 且包含 α 的区间。
(3)
(ii) g(x)=3sinx−3cosx
其中 x 以弧度为单位。
(a) 证明方程 g(x)=0 在区间 [4,5] 内有根 β
(2)
(b) 在区间 [4,5] 上使用一次线性插值法求 β 的近似值。
答案精确到 4 位有效数字。
(2)
解答
(i)
解法一
思路
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从已知区间 [0,1] 开始二分。先计算中点 0.5 的函数值,根据端点异号把区间缩为 [0,0.5];再计算新中点 0.25 的函数值,得到宽度恰为 0.25 的含根区间。必须写出函数值,不能只写正负号。
答题过程
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For
f(x)=x2+5−85x,
we have
f(0)=f(0.5)==4,(0.5)2+5−82.5−175.769…
Since f(0)>0 and f(0.5)<0, the root lies in [0,0.5]. Bisecting this interval again,
f(0.25)==(0.25)2+5−81.25−8.39184…
As f(0)>0 and f(0.25)<0, and f is continuous, the required interval is
[0,0.25].
(ii)(a)
解法一
思路
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分别计算区间两端的函数值。因为指数函数、正弦和余弦都连续,所以 g 在 [4,5] 上连续;端点函数值异号后,使用介值定理即可证明区间内有根。
答题过程
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Using radians,
g(4)==3sin4−3cos42.396…>0,
whereas
g(5)==3sin5−3cos5−0.502…<0.
The function g is continuous on [4,5]. Since g(4) and g(5) have opposite signs, the intermediate value theorem shows that g(x)=0 has a root β in
[4,5].
(ii)(b)
解法一
思路
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承接 (a) 的两个端点函数值,用连接 (4,g(4)) 与 (5,g(5)) 的弦和 x 轴的交点近似根。由于右端函数值为负数,分母必须保留减去负数的符号。
答题过程
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Using linear interpolation once,
β≈==4+g(5)−g(4)0−g(4)(5−4)4+−0.502…−2.396…−2.396…4.826…
Therefore, to 4 significant figures,
β≈4.827.