Determining the number of inflection points, given f'(x) = sin(x^3) for -1.8 <= x <= 1.8

[ausmathgenius420]

The question is as follows: The derivative of a function [imath]f(x)[/imath] is given by [imath]f'(x)=sin(x^3)[/imath] for the domain[imath][-1.8\leq x \leq 1.8][/imath]. Determine the number of inflection points that [imath]f(x)[/imath] has.

I found that [imath]f''(x)=3x^2cos(x^3)[/imath]. When I graph that there is five times that [imath]f''(x)=0[/imath]. When graphing [imath]f'(x)[/imath] I see that [imath]x=0[/imath] is a stationary inflection point. The answer is four but I'm unsure why?

 

[Steven G]

I think that you have the wrong definition for finding a point of inflection. Can you please tell us the definition which you are using?

 

[Steven G]

Consider f(x) = x³
What is f'(x) and when does it equal 0?
What is f"(x) and when does it equal 0?
Is there an extrema at x=0?
Is there a point of inflection at x=0?

Consider f(x) = x⁴
What is f'(x) and when does it equal 0?
What is f"(x) and when does it equal 0?
Is there an extrema at x=0?
Is there a point of inflection at x=0?

Look at these two functions carefully and then settle in your mind how to tell when a function has an extremum and when a function has a point of inflection.