Investigate the properties of the graph of f(t) = sin(3*x)^2 on [0,2/3*Pi]?

Question

Find the sub-intervals in which the function is increasing and decreasing.
Find the sub-intervals in which the function is concave up and down.
Find the inflection points of the function f on the interval? Your answer should be a set of entries of the form inflection_pt(x1,y1)

Answer 1

This is all about determining derivatives and determining their roots.

Since f(t) is continuous, it can only go from increasing to decreasing, or vice versa, at a point where its derivative f'(t) is either 0 or doesn't exist.

If f'(t) is continuous, it can only change from concave up to concave down, or vice versa, at a point where the second derivative, f''(t) is either 0 or doesn't exist.

The points of inflection are the points at which the curvature changes sign.

So the first thing to do is to compute the first and second derivatives of f(t).

Then compute the roots of both derivatives (the points where they are 0) in the interval [0, (2/3)*pi]

Then determine which of these points are really the ones you care about.

(Just because the derivative has to be 0 where a function goes from increasing to decreasing, or vice versa, doesn't mean that all points where the derivative is 0 signal such a shift.

Similarly, while f''(t) = 0 is required for a point to be an inflection point, f''(t) can be 0 without the point being an inflection point.)

When computing the derivatives, you can just slog through the various stages or you can take a shortcut with the double angle formulas:

http://en.wikipedia.org/wiki/List_of_trigonometric_identities

In particular:
sin^2(u) = (1 - cos(2u))/2 so
f(t) = (1 - cos(6t))/2

This makes taking the derivatives easier and finding their roots trivial.