Calculus min and max within an interval
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Calculus min and max within an interval

[From: ] [author: ] [Date: 11-11-12] [Hit: ]
6 . Please help!f (x) = 2.94cos(.6x) + 4.8 sin(.......
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!

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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!

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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.
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