Question

I could really use a walk through of the method for solving this
Approximate the critical number of f(x) = 18x cos (x) on the interval (0,pi) round to three decimal places.

Answer 1

first we need the derivative, it is
\[18(\cos(x)-x\sin(x))\] by the product rule

Answer 2

oh good I managed to get that far lol but that was pretty much it.

Answer 3

critical point will be the numbers in the interval \((0,\pi)\) where
\[\cos(x)-x\sin(x)=0\]

Answer 4

there is no easy way to solve this

Answer 5

you can try
\[\cos(x)=x\sin(x)\] or
\[\frac{\cos(x)}{x}=\sin(x)\] but algebra will not get it for you, you need some sort of technology or newton-ralphson method.
i would use technology

Answer 6

http://www.wolframalpha.com/input/?i=cos%28x%29-xsin%28x%29%3D0

Answer 7

the ugly decimals will show you that there is no snap way to do it

Answer 8

we were working on newton's method I ended up with 1.57 as an approximation of the zero but I was not sure if that was right.

Answer 9

i guess not since that is neither of the answers from wolf

Answer 10

even newton's method will require technolgy because there is no way for you know know what say
\[\sin(1)\] is so forget that mess and cheat

Answer 11

lol ok thanks I did get .86 with my calculator initially.

Answer 12

thank you :)

Answer 13

good, others are there as well

Answer 14

yw

Answer 15

the others aren't in my interval.

Answer 16

really interesting...

Answer 17

yeah only one is in the inteval