find the real root of x=3e^-x by regula falsi method.
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How do l solve 3x=e^x by bisection method?
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Souvik Chakraborty
Answered July 10, 2017
Well, consider the function f(x)=ex−3x
Since e x is an infinite series, the given equation is a transcendental equation which can be solved by various appropriate iterative methods, one of them being the bisection method.
We need to find a domain of x where f(x) changes sign.
Simply by observing the function, we find that f(1)<0 and f(2)>0
that is, f(1)∗f(2)<0
This makes your new domain of interest as (1,2).
At every stage of the subsequent iteration, we will look to find a domain where the function changes sign.
That is, one of the roots of the equation is guaranteed to lie in (1,2).
Now, consider the midpoint of this domain, viz, 1.5.
Find the sign of f(1.5).
Its evident from the substitution that f(1.5)<0.
Thus, our new interval of interest becomes (1.5,2) since f(1.5)∗f(2)<0
Then, use the midpoint (2+1.5)/2=1.75 as the new bisection point and compute the value of f(x) at 1.75 , and thereby, obtain the new interval (a,b) such that f(a)∗f(b)<0.
In this way, after a large number of iterations, we can obtain a sufficiently small interval so that the root converges to the midpoint of that interval, i.e. in that case, for an interval (a,b),|b−a|<ε , where ε is the permissible error.
Hope that gives you the idea on how to solve the given function for its roots.
The more number of iterations you perform for the given transcendental function, the more will you approach an error-free solution, and with every iteration, the value of the root obtained by iteration approaches the actual root.
Here’s the graph of the function f(x)=ex−3x.
And it has multiple roots, hence you will be able to determine other intervals (besides (1,2) ) where the function changes sign, one of the intervals being (0,1).
Step-by-step explanation:
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