f(x) = 3 sin x + 3 cos x 0 ≤ x ≤ 2π. Find the interval in which f is concave up and concave down. Find the inflection points

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f(x) = 3 sin x + 3 cos x 0 ≤ x ≤ 2π. Find the interval in which f is concave up and concave down. Find the inflection points

Mathematics
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First find derivative
3cosx-3sinx
to find inflection points, just take the second derivative, set it to zero and solve for x.

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so the second derivative is -3sinx-3cos x
but how would i solve for that
factor out the -3?
set to 0 solve for x
to find the inflection points, you have to derive your function twice, and then you'll have to take 2 conditions. 1) f''(x) = 0 2) f''(x) = UND depends on the function ofcourse when you set your derived function to zero, you'll get numbers, those are your x's and to find the y's to finish up your process , plug in the x you found in the original function, and with that, you'l have your inflection points ^_^ I hinted it for you, now give it a try :)
what about the concave up and down?
for that, you have to draw a table putting in the x, f''(x), and f in the first column like this : x | (--) 0 (+) ___|________________________ f'' | ___|_________________________ f | then you take negative values for x and plug them in the original function, if it gives you a negative answer then it's concaved down, and if it gives you a positive answer then it's concaved up. Same story goes when you take positive values for x~ after that, from the table, you'll be able to determine the intervals of concavity :)
This one is tricky. I think you have to plot sin x - cos x over your interveral 0 to 2 pi. You would get your max and min because it is a periodic function
oh, right, so you'll hav to list the x's on the x's row lol, you know -2pi -pi -pi/2 0 pi/2 ...e.tc :)
but since its between 0 and 2pi i do 0, pi/2, pi/3, pi?
Been a while since I did this. We need Lokisan or Xavier
You make a table x=0, f' (x) (or f''(x) which ever one=sinx - cos x. You evaluate over 4 intervals, you would get your max, min

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