Please tell me how to do this.
Use Newton's method to find the absolute maximum value of the function f(x) = xcos(x), 0 ≤ x ≤ , correct to six decimal places.
it seems to me i've seen this problem before. you're not double posting, are you?
anyway, here goes. i'll start you off.
so remember, for the absolute max, we want to find the largest value for the function on the interval we are considering. now this value may occur at the endpoints, or it may occur at a local maximum within the interval, so let's check those.
$\displaystyle f(x) = x \cos x$ for $\displaystyle 0 \le x \le \pi$
$\displaystyle \Rightarrow f'(x) = \cos x - x \sin x$, $\displaystyle 0 \le x \le \pi$
(1) check the value of the function at the end points
$\displaystyle f(0) = 0$
and
$\displaystyle f(\pi) = - \pi$
(2) check the value of the function at the critical points
for the critical points, set $\displaystyle f'(x) = 0$
$\displaystyle \Rightarrow \cos x - x \sin x = 0$
now this is relatively hard to solve, here is where we will need Newton's method, to find the critical point.
recall that Newton's method says that if we are given an approximation $\displaystyle x_n$ for the zero of a function, we can get a better approximation $\displaystyle x_{n + 1}$ by way of the formula:
$\displaystyle x_{n + 1} = x_n - \frac {f(x_n)}{f'(x_n)}$ .....................(1)
now, take your first guess to be 1. thus you have $\displaystyle x_1 = 1$
so you will find the second approximation, $\displaystyle x_2$, by:
$\displaystyle x_2 = 1 - \frac {f(1)}{f'(1)}$
now calculate $\displaystyle x_2$. when you get $\displaystyle x_2$, plug it in for $\displaystyle x_n$ in (1) to get the third approximation $\displaystyle x_3$, that is, find:
$\displaystyle x_3 = x_2 - \frac {f(x_2)}{f'(x_2)}$
continue to do this until you realize that your value for $\displaystyle x_n$ has the first 6 decimal places constant. then take that to be the critical point. then, calculate the value of the function at that point. then pick the absolute max.
(note, think of $\displaystyle x$ to be in radians)