At the roots,
x/6 - 6sec x = 0
x/6 = 6secx
x = 36secx
There is no simple algebraic way to solve an equation like this, so
we need to use "technology"
We can get an approximate answer by graphing
y = x and y = 36secx
I used
www.desmos.com/calculator
but had to zoom out a bit to see y = 36sec x
It showed the first at x = 37.423 and a 2nd at x = 38.027
the next one is beyond our domain.
Find all the roots of the given function on the given interval. Use preliminary analysis and graphing to find good initial approximations.
f(x) = x/6 - 6 sec (x) on [0,40]
1. The function has root(s) when x =
3 answers
well, just looking at the graph, f(x) has two roots on [0,40].
Since f(x) has asymptotes at odd multiples of π/2, I'd start just to the right of 23(π/2) and to the left of 25(π/2) or, at, say 37 and 39.
Then, Newton's method will be sure to converge on the roots (why?).
Starting with x = 37, iterations produce
37, 37.2475, 37.3829, 37.4199, 37.4225
Now see what you can do with x=39
Since f(x) has asymptotes at odd multiples of π/2, I'd start just to the right of 23(π/2) and to the left of 25(π/2) or, at, say 37 and 39.
Then, Newton's method will be sure to converge on the roots (why?).
Starting with x = 37, iterations produce
37, 37.2475, 37.3829, 37.4199, 37.4225
Now see what you can do with x=39
ALL OF U R WRONG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! LO$ER$
2 answers