Asked by Kaur
Find the intervals on which sinx-root3(cosx) is increasing and decreasing
Very urgent
Very urgent
Answers
Answered by
oobleck
does root3(cosx) mean cube root? If so, then
y = sinx - (cosx)^(1/3)
y is increasing when y' > 0
y' = cosx - 1/3 (cosx)^(-2/3) * (-sinx)
The graph is at
https://www.desmos.com/calculator/nwct6fs2np
I'm surprised at the abrupt changes of direction, since cosx is smooth at odd multiples of pi/2.
The graph of y' is at
https://www.wolframalpha.com/input/?i=cosx+-+1%2F3+%28cosx%29%5E%28-2%2F3%29+*+%28-sinx%29
you can see that y'=0 once every 2pi. y is decreasing/increasing in between the zeroes of y'
y = sinx - (cosx)^(1/3)
y is increasing when y' > 0
y' = cosx - 1/3 (cosx)^(-2/3) * (-sinx)
The graph is at
https://www.desmos.com/calculator/nwct6fs2np
I'm surprised at the abrupt changes of direction, since cosx is smooth at odd multiples of pi/2.
The graph of y' is at
https://www.wolframalpha.com/input/?i=cosx+-+1%2F3+%28cosx%29%5E%28-2%2F3%29+*+%28-sinx%29
you can see that y'=0 once every 2pi. y is decreasing/increasing in between the zeroes of y'
Answered by
Kaur
No its not cube root
Sinx - (root3)(cosx)
Sinx - (root3)(cosx)
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