<b>(a)</b> f(r)=f(x)+f'(x)(r-x)+f''(x)(r-x)<sup>2</sup>/2+ ... <b>(b)</b> r=x-f(x)/f'(x) <b>(c)</b> f'(x)=(f(x+h)-f(x-h))/2h <b>(d)</b> r=x-2 h f(x)/(f(x+h)-f(x-h)) <b>(e)</b> f'(x)=(f(x+h)-2 f(x)+f(x-h))/h<sup>2</sup> <b>(f)</b> f''(x)(r-x)<sup>2</sup>/2+f'(x)(r-x)+f(x)=0 <b>(g)</b> r=x-(f'(x)+([f'(x)]<sup>2</sup>-2 f''(x)f(x)))/f''(x)
Example 1: Newton's method is based on the Taylor expansion.