Question

curve sketching problem
y=sin^(3)x

Answer 1

i got the domain and intercepts. i need help in finding the symmetry and asymptotes

Answer 2

No asymptotes to worry about - \(\sin^3x\) is defined for all \(x\), and oscillates as \(x\to\pm\infty\).

For symmetry, you'll need to check if it's even, odd, or neither.
\[\sin^3(-x)=-\sin^3x~~\implies~~\text{odd}\\
\sin^3(-x)=\sin^3x~~~~~~\implies~~\text{even}\]

Answer 3

Also, I remember curve sketching having a lot to do with finding critical points/extrema and inflection points. Are you supposed to do that too?

Answer 4

its not negative its positive sin^3x

Answer 5

so its going to be neither right? for the symmetry?

Answer 6

@SithsAndGiggles

Answer 7

Yes, I'm aware that \(\sin^3x\) is your given function, but to check for symmetry, you have to see what happens for negative arguments \(-x\) as opposed to any argument \(x\).
\[\sin^3(-x)=(\sin (-x))^3=(-\sin x)^3=-(\sin x)^3=-\sin^3x\]
and so \(\sin^3x\) is odd.