Question

Finding critical point of 6cosx-2x=0?

Answer 1

how do i go about this?

Answer 2

why in physics section? -_-

Answer 3

haha. its 5am, excuseee me

Answer 4

anyways, I excuse you.
f'(x) = 0
what do you get ?

Answer 5

well the initial equation is 6sinx-x^2. I did the first derivative and got 6cosx-2x=0 and now i need to find the critical points from it.

Answer 6

hmm, okay
so then ,
3 cosx = x

seems complicated this one.

Answer 7

the question must be asking about the "number" of critical points, isn't it?

Answer 8

it is very! and yes!

Answer 9

so did you factor this?

Answer 10

well if its the no. of critical points, then it ain't much tough :|

Answer 11

here you don't have to find exact values of x which satify this eqn.
but you have to find no. of "x" that satisfy this.

Answer 12

umm its asking for any critical numbers.

Answer 13

must be only what I am assuming, otherwise its very tough, will then require the knowledge of advanced mathematics, namely interpolation.

Answer 14

yes i used interpolation for finding the x intercept

Answer 15

omg, you have studied interpolation? :O
I haven't yet! -_-

Answer 16

i dont believe it is f(x)=6sin(x-x^2) because it says to round any answers to two decimal places.

Answer 17

I dont think so! lol

Answer 18

its pretty crazy looking

Answer 19

-_-
question is f(x) = sin(x-x^2) and not sinx - x^2 ?

Answer 20

Im not sure. there is no parenthesis in the equation.

Answer 21

But i havent studied interpolation so, it doesnt make sense for it to be assigned.

Answer 22

Pretty confusing.

Answer 23

probably you should ask this in maths section then,
some experts out there will surely help you out :)
ask the original ques there

Answer 24

thanks :) I asked earlier about the x intercept but i will wait and ask my teacher about it, no sense in guessing and wasting time on the incorrect equation. thankk youu so much