Question

Question #fb316

Answer 1

Keep up the good work

Explanation:

You have started out well, although it would have been easier to follow your line of thought if you wrote that the `x`-values you found are the `x` of your stationary points (points where (`y'=0`)).

To continue, you can reason as follows:

1. Your function is continous and differentiable, so the only places where the solution could change from increasing to decreasing or vice versa are stationary points.

2. It is important to remember that not all stationary points where `(y'=0)` are extrema, since they could also be saddle points.

3. To check whether your stationary points are extrema or not, you can check if the second derivative is non-zero or not. Another option is to check the sign of the first derivative `y'` at one point each in each interval between a stationary point and/or endpoint.

4. If you choose the second derivative test, then `y''<0` means that the stationary point is a maximum. Consequently, the function is rising to the left of the maximum, and decreasing to the right of the maximum. If `y''>0` you have a minimum, and you can figure out yourself on which sides the function is increasing and decreasing. If `y''=0` you have a saddle point, and you need to look at values around the stationary point to see if the function is increasing or decreasing around the stationary point.

5. If you choose to evaluate `y'` at a set of points instead, then `x = pi/2`, `x = 9 pi/12` and `x=pi` are good candidates (convince yourself why).

6. As for evaluating for instance `sin((7 pi)/6)`, recall the trigonometric identity `sin(x + pi) = -sin(x)`, from which it follows that `sin((7 pi)/6) = sin((6pi)/6 + pi/6) = sin(pi + pi/6) = -sin(pi/6)`.

Good luck! Feel free to write if you need more tips.