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

MATH104 December 2010

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[hide] Question 01 (g)
You are using the Newton-Raphson Method to approximate a solution of an equation f(x) = x and you make an initial guess $x_{0}=3$ to the solution. If the tangent line to y = f(x) at x = 3 has the equation y = 5x - 7, what is the next approximation $x_{1}$ to the solution?

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If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
Notice that finding a solution to f(x)=x is the same as finding the root of an equation g(x)=f(x)-x.

[show] Hint 2
How can you use the tangent line y=5x-7 to give you information about $\displaystyle {}f(x_{0})$ and $\displaystyle {}f'(x_{0})$ ?

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[show] Solution
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 $x_{1}=7/4$ .