Science:Math Exam Resources/Courses/MATH104/December 2010/Question 01 (g)/Solution 1

We want to find a solution to

$\displaystyle {}f(x)=x$

and so we begin by defining a function

$\displaystyle {}g(x)=f(x)-x$

so that we can think instead of finding roots to g(x). Recall that the Newton-Raphson Method for finding roots to a function g(x) gives a future approximation ($x_{n+1}$) based on a current approximation ($x_{n}$) as

$x_{n+1}=x_{n}-{\frac {g(x_{n})}{g'(x_{n})}}.$

Therefore, in our case

$x_{1}=x_{0}-{\frac {g(x_{0})}{g'(x_{0})}}=3-{\frac {g(3)}{g'(3)}}$

since $x_{0}=3$. The task now is to find what g(3) and g'(3) are. Since g(x)=f(x)-x then

$\displaystyle {}g'(x)=f'(x)-1$

so we really only need to determine f(3) and f'(3). To do this we can use the tangent line which is also the linear approximation to f(x). Therefore we have that

$\displaystyle {}f(x)\approx {}5x-7=5x-15+8=8+5(x-3)=f(3)+f'(3)(x-3).$

Therefore, we have that f(3)=8 and f'(3)=5. Similarly, we then have that,

${\begin{aligned}g(3)=f(3)-3=8-3&=5\\g'(3)=f'(3)-1=5-1&=4\end{aligned}}$

and so we get that,

$x_{1}=3-{\frac {g(3)}{g'(3)}}=3-{\frac {5}{4}}={\frac {7}{4}}.$

Therefore the next approximation to the solution is $x_{1}=7/4$.