Give brainliest if you tell me the can tell me the most random thing
1 answer:
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Answer:
The root is:
Explanation:
Use this script in Matlab:
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function [c, err, yc] = bisect (f, a, b, delta)
% f the function introduce as n anonymous function
% - a y b are the initial and the final value respectively
% - delta is the tolerance or error.
% - c is the root
% - yc = f(c)
% - err is the stimated error for c
ya = feval(f, a);
yb = feval(f, b);
if ya*yb > 0, return, end
max1 = 1 + round((log(b-a) - log(delta)) / log(2));
for k = 1:max1
c = (a + b) / 2;
yc = feval(f, c);
if yc == 0
a = c;
b = c;
elseif yb*yc > 0
b = c;
yb = yc;
else
a = c;
ya = yc;
end
if b-a < delta, break, end
end
c = (a + b) / 2;
err = abs(b - a);
yc = feval(f, c);
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Enter the function in matlab like this:
f= @(x) 0.005*(exp(2*x)*cos(x))
You should get this result:
f =
function_handle with value:
@(x)0.005*(exp(2*x)*cos(x))
Now run the code like this:
[c, err, yc] = bisect (f, 1, 2, 1e-10)
You should get this result:
c =
1.5708
err =
5.8208e-11
yc =
-3.0708e-12
In addition, you can use the plot function to verify your results:
fplot(f,[1,2])
grid on
