f(x) = cosx+x-2
f'(x) = 1-sinx
start with, say, x=3
x -> 3 - f(3)/f'(3) = 3-.01/.8588 = 2.98835
and proceed as needed
Approximate the real root of cosx+x=2 to four decimal place. (Using Newton's method)
I need help with this! Thank you
2 answers
I will start you ought. A neat way to get to near the solution is to graph
cosx + x -2
look at where the graph crosses the x axis. Then change the limits on the graph to expand the area just around that intercept. Keep doing that until you can read it to 5 significant figures....
That is faster than the old method of using the calculator (non grahic) to iterate until you get the solution.
cosx + x -2
look at where the graph crosses the x axis. Then change the limits on the graph to expand the area just around that intercept. Keep doing that until you can read it to 5 significant figures....
That is faster than the old method of using the calculator (non grahic) to iterate until you get the solution.
1 answer