What does the following code do?

myFunction =@(x)x^2-6;

x_lower=0;

x_upper=5;

x_mid=(x_lower+x_upper)/2;

while abs(myFunction(x_mid))>0.01

if (myFunction(x_mid)*myFunction(x_upper))<0

x_lower =x_mid;

else

x_upper = x_mid;

x_mid=(x_lower+x_upper)/2;

end

x_mid=(x_lower+x_upper)/2;

end

fprintf('the root is %g; x_mid)

[KALYAN ACHARJYA]

Which line do you have issue?

[Raban Nghidinwa]

This is a numerical methods question which requires us to explain what the code does for 16 marks.

[Star Strider]

So asking us to explain it to you is doing your homework for you, giving you an unfair advantage over your classmates who are doing this themselves, likely without any outside help.

[Raban Nghidinwa]

Mr Strider,I am new to MTLAB and we have limited resources but we are tested on an international scale so its struggle sometimes,but thank you so much.

Answer 1

[KALYAN ACHARJYA]

Here myFunction evaluates the value based on x

myFunction=@(x)x^2-6;

Say

myFunction(2), here x=2 passes to function arguments, so the result will

ans =
    -2

Then

Assigned

x_lower=0;
x_upper=5;
x_mid=(x_lower+x_upper)/2;

While loop continue till Absolute(myFunction(x_mid)) is greater than 0.01

{

Then replace the value of x_mid, x_lower, x_upper based on if condition

If true

x_lower change to x_mid

otherwise

x_upper=x_mid;
x_mid=(x_lower+x_upper)/2;

end if

Re-evaluate x_mid again based on present x_lower and x_upper value

Then print the x_mid value

Note: Text the code with some example, change the my function expression and see the results

Hope it helps!

[Raban Nghidinwa]

It helps,thank you.