Let x=√12⇒x2=12 ⇒‌‌x2−12=0 ⇒‌‌f(x)=x2−12 On differentiating w.r.t., x, we get ‌f′(x)=2x ‌∵‌‌f(3)<0‌ and ‌f(4)>0 ‌ Hence, root will lie between 3 and 4 . ‌|f(3)|<|f(4)| ‌∴‌‌x0=3 First iteration, x1‌=x0−‌