Question

MEDALS

Answer 1

I haven't used a graphing calculator in years, but I guess you can just find the derivative and set it to 0 and solve.

Answer 2

and that would be e^-sin(x)

Answer 3

Wrong, remember the chain rule?

Answer 4

And the derivative of e^x?

Answer 5

is it e^2?

Answer 6

No, the derivative is \[-sinx(e^{cosx})\]

Answer 7

ohh ok

Answer 8

I'm not sure if there are even roots for this problem

Answer 9

I guess you'll need a calculator for the rest of it, try looking for a graphing calculator online, prob find one.

Answer 10

found one @iambatman https://www.desmos.com/calculator

Answer 11

is it 3? @aum

Answer 12

@jdoe0001 PLEASE

Answer 13

just find when the following is true on the unit circle
\[\sin(x)=0 \]

Answer 14

since e^cos(x)) will never be 0
the only factor that can be 0 is the sin(x) factor

Answer 15

so it would be 2

Answer 16

http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiItc2luKHgpKmVeKGNvcyh4KSkiLCJjb2xvciI6IiNENjI3MjcifSx7InR5cGUiOjEwMDB9XQ-- you can zoom in/out using the middle-mouse button, and you can also drag with the mouse


if you need a graphing calculator http://www.graphcalc.com/download.shtml
or you can also use https://www.desmos.com/calculator

Answer 17

Yeah I figured there were no solutions, but use the graphing calculator as it will probably give you a better answer.

Answer 18

yes @mondona one at x=0 and one at x=2pi
But I guess that is actually without a graphing calc

Answer 19

If you wanted to do it just with your graphing calculator..you don't even really need to find f'

Answer 20

thank you(:

Answer 21

Just graph the original function f and see where you have horizontal tangents from x=0 to x=2pi including at those numbers