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

MATH104 December 2010

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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?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

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.

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.

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

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $x_{1}=7/4$.

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