Question

Is it even possible to graph sketch this?:

f(x) = x^2sin(4x)?

When I take the derivative, I get:
f'(x) = 2x*sin(4x) + x^2*cos(4x)4
0 = 2x*sin(4x) + x^2*cos(4x)4
0 = 2x(sin4x + 2x*cos(4x))

I know for sure that x = 0 is a critical point. However, I am not sure how to find x for the second bracket.

Answer 1

I am trying to get x = _ for that second bracket for f', but I am having trouble doing so.

Answer 2

i wouldn't use calculus

Answer 3

sine is bounded by -1 and 1
so x^2 sine is bounded by x^2 and -x^2

Answer 4

OHHHH