Calculus min and max within an interval

I'm having a lot of trouble with this particular problem..
http://i43.tinypic.com/1iif0g.png

I think the derivative is arctan(4.9/8)/.6, but whenever I plug it into the original equation, the computer tells me it isn't correct. Also, I am very confused about the min and max values thing, and you can tell the difference when the derivative is just arctan(4.9/8)/.6 . Please help!

You may have just a sign error:

f ' (x) = 2.94cos(.6x) + 4.8 sin(.6x)=0

Then 2.94/4.8= -sin(.6x)/cos(.6x)

-.6125 = tan(.6x)

So you are on the right track. The endpoints of the interval are the zeroes (or very close)

solve [arctan(-.6125)] = -.54956...+/- Kpi= -.550, 2.590, -3.69, 5.73, etc

Then divide by 0.6 to find the answers that are in (3.53,6.93)

X= -.92 or 4.32

Find f(-.92) and f(4.32) then compare to the endpoints (zero).

If you graph the function on your calculator, this will help you see the critical points.

Hoping this helps!

It looks like you set the function equal to zero then solved for x to get arctan(4.9/8) / 0.6
This will only tell you where x-intercepts are. You need to find the derivative first.

f ' (x) = 2.94 cos(0.6x) + 4.8 sin(0.6x)

Set the derivative equal to zero and solve for x to get critical point(s).
Evaluate f(x) at the critical point(s) and endpoints. The greatest of these numbers will be the max. and the least will be the min.