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draveon353
29.11.2020 •
Mathematics
Consider the following.
cos(x) = x3
A) Prove that the equation has at least one real root.
The equation cos(x) = x3 is equivalent to the equation f(x) = cos(x) − x3 = 0. f(x) is continuous on the interval (0, 1) there is a number c in (0, 1) such that f(c) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation cos(x) = x3, in the interval (0, 1).
B) Use your calculator to find an interval of length 0.01 that contains a root.
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Ответ:
Step-by-step explanation:
Given that:
cos(x) = x³
f(x) = cos (x) - x³
f(x) is continuous on the interval (0, 1)
when;
f(0) = cos (0) - (0)
f(0) = 1 - 0
f(0) = 1 > 0
f(1) = cos (1) - 1³
f(1) = 0.5403 - 1
f(1) = -0.4597
f(1) = - 0.46
f(1) = - 0.46 < 0
Since, 1 > 0 > -0.46, thus there is a number ''c" in (0,1)
such that f(c) = 0
By applying the Intermediate Value Theorem, there is a root of the equation in the interval (0,1)
b) Given that:
The interval length = 0.01; this implies that it is 0.005 length from its root.
f(0.865) = 0.0014
f(0.87) = - 0.013
The solution to the interval lies between 0.865 , 0.87 by using a calculator at length 0.01.
Ответ:
1=8.25
2=16.5
3=24.75
no
Step-by-step explanation:
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