Question

The derivative of the function f is given by
f ' (x)=x^2 (cos(x^2)). How many points of inflection does the graph of f have on the open interval (-2, 2)?

Answer 1

equating it to zero
we have x^2 cos(x^2) = 0

so obviuosly x = 0 is one solution

Answer 2

I thought I had to find the second derivative?

Answer 3

yes you have to

Answer 4

in order to find the nature of the turning points

Answer 5

i was just saying that there is one turning point at x = 0

Answer 6

but there could be more between -2 and + 2

Answer 7

by the zero product law cos (x^2) = 0 as well

Answer 8

ok let see if I got the second derivative correct. I think they meant for me to use a graphing calculator.
f " (x)=2x (cos(x^2))-sin(x2)(2x)x^2

Answer 9

f " (x)=2x (cosx^2-x^2sinx^2)

Answer 10

thanks I need to go look up some more things to help me with this concept

Answer 11

yea looks good

Answer 12

well you need to find all the zeros for x^2 cos (x^2) first

Answer 13

the cosine is zero for pi/2 and - pi/2

Answer 14

@mathmale

Answer 15

solution is three but I am interested in the process and how it is three

Answer 16

@zepdrix