anonymous
  • anonymous
NEED HELP! In finding the absolute extrema of the function on the given interval of [pi/4, 7pi/4] when g(x) = x + cot(x/2)
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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freckles
  • freckles
So first step is to differentiate to find critical numbers.
anonymous
  • anonymous
wait, whats the derivative of cot?
freckles
  • freckles
the derivative of cot(x) w.r.t. x is -csc^2(x).

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anonymous
  • anonymous
oh i know
freckles
  • freckles
you have to use chain rule there since you have x/2 inside though
freckles
  • freckles
\[\frac{d}{dx}\cot(u(x))=u'(x) \cdot [-\csc^2(u(x))] \\ \frac{d}{dx}\cot(u(x))=-u'(x) \cdot \csc^2(u(x))\]
anonymous
  • anonymous
yea I was about to type that... so basically the derivative is g'(x) = 1+ -csc^2(x/2) * 1/2
freckles
  • freckles
yeah and we set that =0 to find critical numbers
freckles
  • freckles
\[1-\csc^2(\frac{x}{2}) \cdot \frac{1}{2} =0 \\ 2 -\csc^2(\frac{x}{2})=0 \\ \csc^2(\frac{x}{2})=2 \\ \sin^2(\frac{x}{2})=\frac{1}{2}\] that should make it easier for you to solve that last step was just taking reciprocal of both sides
anonymous
  • anonymous
wait...how did you get it to become sin^2 (x/2) = 1/2 ?
freckles
  • freckles
well first step I multiplied 2 on both sides
freckles
  • freckles
second step I added csc^2(x/2) on both sides
anonymous
  • anonymous
I got everything until the last part
freckles
  • freckles
last step I took the reciprocal of both sides
anonymous
  • anonymous
so you did: (1/2)*(2) = 1/csc*csc^2(x/2)
freckles
  • freckles
I don't think that is what I did.
freckles
  • freckles
I'm not sure what that is.
anonymous
  • anonymous
lol could you show me the work for how you got what you got in the last part
freckles
  • freckles
\[\text{ if } \frac{1}{a}=\frac{1}{b} \text{ then } a=b\]
anonymous
  • anonymous
yea thats true
freckles
  • freckles
\[\csc^2(\frac{x}{2})=2 \\ \frac{1}{\sin^2(\frac{x}{2})}=2 \\ \frac{1}{\sin^2(\frac{x}{2})}= \frac{1}{\frac{1}{2}} \\ \text{ which means } \sin^2(\frac{x}{2})=\frac{1}{2}\]
anonymous
  • anonymous
oh you just switch out csc to 1/sin
anonymous
  • anonymous
k i get it
freckles
  • freckles
yes those are reciprocal functions
freckles
  • freckles
csc(u) is the reciprocal sin(u) and vice versa
freckles
  • freckles
which means csc(u)=1/sin(u)
anonymous
  • anonymous
so now we find the critical points.. if its sin(x/2)^2 = 1/2
anonymous
  • anonymous
then would it be pi/6?
anonymous
  • anonymous
Oh wait, that doesnt fit the interval
freckles
  • freckles
\[\sin^2(\frac{x}{2})=\frac{1}{2} \implies \sin(\frac{x}{2})=\frac{\sqrt{2}}{2} \text{ or } \sin(\frac{x}{2})=-\frac{\sqrt{2}}{2}\]
anonymous
  • anonymous
how did you get a sqrt of 2 in the second part, wasnt there a 1 there before?
freckles
  • freckles
\[\sqrt{\frac{1}{2}}=\frac{\sqrt{1}}{\sqrt{2}}=\frac{1}{\sqrt{2}} =\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2} \\ \]
anonymous
  • anonymous
oh okay
freckles
  • freckles
i didn't do all those steps I just remember that sqrt(1/2) =1/sqrt(2) or I also know I can write sqrt(2)/2
anonymous
  • anonymous
Yea thats a good thing to keep a note of
anonymous
  • anonymous
wait... how do you solve for that x...? I was assuming it could be pi/4, 3pi/4...but it just doesnt look right with how the x is placed in the equation
freckles
  • freckles
\[\text{ we have two equations \to solve } \text{ Let } u=\frac{x}{2} \\ \sin(u)=\frac{\sqrt{2}}{2} \text{ or } \sin(u)=\frac{-\sqrt{2}}{2} \\ u=\frac{\pi}{4}+2n \pi , \frac{-\pi}{4}+ 2n \pi \text{ or } u=\frac{2 \pi}{3}+2 n \pi , \frac{5\pi}{4}+2 \pi n \\ \text{ or condensing this a little } \\ u=\frac{\pi}{4}+\pi n \text{ or } \frac{-\pi}{4} + n \pi \\ \text{ now remember we wanted } \frac{\pi}{4} \le x \le \frac{7 \pi}{4} \\ \text{ but we solved for } u \\ \\ \text{ recall } u=\frac{x}{2} \\ \text{ so } x=2u \\ \text{ so we need to find } u \text{ above such that } \\ \frac{\pi}{4} \le 2u \le \frac{ 7\pi}{4} \\ \text{ dividing 2 on all sides gives } \\ \frac{\pi}{8} \le u \le \frac{ 7\pi}{8} \] so before we go further and solve for x can you solve for u on the interval [pi/8,7pi/8]
anonymous
  • anonymous
wow, didnt think of splitting the equation into 2 separate parts. But yell ill solve for u on the interval
anonymous
  • anonymous
I dont know how to find where pi/8 and the other value is onthe unit circle
freckles
  • freckles
\[\text{ we are solving } \sin(u)=\frac{\sqrt{2}}{2} \text{ or } \sin(u)=\frac{-\sqrt{2}}{2} \text{ on } u \in [\frac{\pi}{8},\frac{7\pi}{8}] \\ u=\frac{\pi}{4}, \frac{3\pi}{4}\]
freckles
  • freckles
now since u=x/2 we need to solve: \[\frac{x}{2}=\frac{\pi}{4} \text{ and also solve } \frac{x}{2}=\frac{3\pi}{4}\]
anonymous
  • anonymous
so x= 2pi/4 and 6pi/4
freckles
  • freckles
reducing x=pi/2 or x=3pi/2
freckles
  • freckles
now you have critical points on the given interval so plug in the critical numbers and the endpoints into f to see which gives you highest output and which gives you lowest output
anonymous
  • anonymous
do i plug the x values in the derivative or the original function?
freckles
  • freckles
also for the above equation if it would have made things easy we could have also used the half angle identity for sin \[\sin^2(\frac{x}{2})=\frac{1}{2} \\ \frac{1}{2}(1-\cos(x))=\frac{1}{2} \\ \text{ multiply both sides by } 2 \\ 1-\cos(x)=1 \\ \text{ subtract 1 on both sides } -\cos(x)=0 \\ \cos(x)=0 \\ \text{ this probably would have been easier for you \to solve }\]
freckles
  • freckles
no you plug into the original function the derivative will only tell you if the slope is positive,0, negative ,or non-existing
freckles
  • freckles
easier in general not just you lol
freckles
  • freckles
\[g(x)=x+\cot(\frac{x}{2}) \\ g(\frac{\pi}{4})=\frac{\pi}{4}+\cot(\frac{\pi}{8}) \\ g(\frac{\pi}{2})=\frac{\pi}{2}+\cot(\frac{\pi}{4}) \\ g(\frac{3\pi}{2})=\frac{3\pi}{2}+\cot(\frac{3\pi}{4}) \\ g(\frac{7\pi}{4})=\frac{7\pi}{4}+\cot(\frac{7 \pi}{8})\]
freckles
  • freckles
so you just might want to whip out your calculator and plug those into see which gives the highest and lowest output
anonymous
  • anonymous
okay, I see
anonymous
  • anonymous
This problem required a lot of work...I thought it wouldve been easy, I guess I have to review the concept behind this problem
freckles
  • freckles
well I think the only hard part was the way we solved for the critical numbers in the beginning the half-identity made the equation faster to solve
freckles
  • freckles
half angle-identity *
freckles
  • freckles
but either way will suffice
anonymous
  • anonymous
got it, thanks!
freckles
  • freckles
cool stuff
freckles
  • freckles
peace
anonymous
  • anonymous
Step1- find critical values and replace them in original function Step2- replace endpoints as well Step3- check your highest and lowest values Done :)

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