Thread: Functions and Newton-Raphson formula

Howdy, the forum looks great.

I have the below question.

http://i.imgur.com/xMJ7fs3.jpg

I'm looking for someone to break down it down for me as I don't have any solved examples to work off. I only have the questions, so having one with a break down should hopefully help me solve the others.

Hope that makes sense.

The first part is easy- f(0)= 0- 3cos(0)= -1 and $\displaystyle f(\pi/2)= \pi/2- 3(cos(\pi/2))= \pi/2$. Since f is negative at 0, positive at $\displaystyle \pi/2$, and f is continuous, it must be 0 some place in between.

As for the second, do not know what the "Newton-Raphson" algorithm is? If your text has this question, surely it at least defines it?

The Newton-Raphson algorithm, to solve the equation f(x)= 0, requires that you define a sequence of x values, $\displaystyle \{x_n\}$, where $\displaystyle x_0$ is some chosen value, hopefully near a root, and then $\displaystyle x_{n+1}= x_n- \frac{f(x_n)}{f'(x_n)}$.