f(x)= cosx/sin^2x+3

I differentiated this equation and got:

f'(x)=[(-sinx)(sin^2x+3)]-[(2sinx)(cosx)(cosx)]/(sin^2x+3)^2

When I simplified this I got

*(1-cos^2x=sin^2x)* NOTES

-sinx(sin^2x+3-2cos^2x) ---> -sinx(1-cos^2x+3-2cos^2x) ----> -sin(cos^2x-2) ---> sinx(-cos^2x+2) --->

sinx=0 ---> x= pi, 2pi. . . nPi

and

cos^2x=2

But, I think that at every multiple of pi there is a V.A

I'm very confused and not sure if I did something wrong. . .

**0**

# Finding critical numbers of Trig functions

Started by Tribalchanter, Oct 27 2011 08:02 PM

1 reply to this topic

### #1

Posted 27 October 2011 - 08:02 PM

### #2

Posted 20 December 2011 - 10:44 PM

well it is a trig function so it will have periodic maxima and minima, your solution looks fine and a plot of the function: http://www.wolframal...2%28x%29%2B3%29

confirms it does infact have alternating maxima and minima at nπ for all integer values for n

confirms it does infact have alternating maxima and minima at nπ for all integer values for n

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