Question

what interval is this increasing on

-3sin^3 x (this is the derivative form)

Answer 1

HI jinnie :)
do some magic :D lol

so it is \[\large f'(x)=-3sin^3x\]
Right friend?

Answer 2

yes

Answer 3

OK remember rules now.. if
\[\large f'(x)<0\] then function is decreasing.
if\[\large f'(x)>0\]then function is increasing.

Answer 4

yea i tried those already

Answer 5

i have tried all these questions im posting. i cant get a result hence the reason im posting them

Answer 6

if you know how to do this, can you just help me out

Answer 7

its a hw question thats due soon and ive tried all i know

Answer 8

@Jinnie is \(-3\sin^3(x)\) the funtion or the derivative of the function?

Answer 9

sine is positive on \((0,\pi)\) and negative on \((\pi,2\pi)\)

Answer 10

therefore \(-3\sin^3(x)\) is negative on \((0,\pi)\) and postive on \((\pi,2\pi)\)

Answer 11

therefore your origninal function, whatever it is, is increasing on \((\pi,2\pi)\) and decreasing on \((0,\pi)\)

Answer 12

thank you very much satellite.

this is the derivative of -3sin^3(x)

-9sin^2(x) cos(x)
what would you say this is conave up and down on and the point of inflection.


lastly for -3sin^3(x)...i got the local min to be pi, cant figure out the local max