Question

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)

Answer 1

So first step is to differentiate to find critical numbers.

Answer 2

wait, whats the derivative of cot?

Answer 3

the derivative of cot(x) w.r.t. x is -csc^2(x).

Answer 4

oh i know

Answer 5

you have to use chain rule there since you have x/2 inside though

Answer 6

\[\frac{d}{dx}\cot(u(x))=u'(x) \cdot [-\csc^2(u(x))] \\ \frac{d}{dx}\cot(u(x))=-u'(x) \cdot \csc^2(u(x))\]

Answer 7

yea I was about to type that... so basically the derivative is g'(x) = 1+ -csc^2(x/2) * 1/2

Answer 8

yeah and we set that =0 to find critical numbers

Answer 9

\[1-\csc^2(\frac{x}{2}) \cdot \frac{1}{2} =0 \\ 2 -\csc^2(\frac{x}{2})=0 \\ \csc^2(\frac{x}{2})=2 \\ \sin^2(\frac{x}{2})=\frac{1}{2}\]
that should make it easier for you to solve
that last step was just taking reciprocal of both sides

Answer 10

wait...how did you get it to become sin^2 (x/2) = 1/2 ?

Answer 11

well first step I multiplied 2 on both sides

Answer 12

second step I added csc^2(x/2) on both sides

Answer 13

I got everything until the last part

Answer 14

last step I took the reciprocal of both sides

Answer 15

so you did: (1/2)*(2) = 1/csc*csc^2(x/2)

Answer 16

I don't think that is what I did.

Answer 17

I'm not sure what that is.

Answer 18

lol could you show me the work for how you got what you got in the last part

Answer 19

\[\text{ if } \frac{1}{a}=\frac{1}{b} \text{ then } a=b\]

Answer 20

yea thats true

Answer 21

\[\csc^2(\frac{x}{2})=2 \\ \frac{1}{\sin^2(\frac{x}{2})}=2 \\ \frac{1}{\sin^2(\frac{x}{2})}= \frac{1}{\frac{1}{2}} \\ \text{ which means } \sin^2(\frac{x}{2})=\frac{1}{2}\]

Answer 22

oh you just switch out csc to 1/sin

Answer 23

k i get it

Answer 24

yes those are reciprocal functions

Answer 25

csc(u) is the reciprocal sin(u)
and vice versa

Answer 26

which means csc(u)=1/sin(u)

Answer 27

so now we find the critical points.. if its sin(x/2)^2 = 1/2

Answer 28

then would it be pi/6?

Answer 29

Oh wait, that doesnt fit the interval

Answer 30

\[\sin^2(\frac{x}{2})=\frac{1}{2} \implies \sin(\frac{x}{2})=\frac{\sqrt{2}}{2} \text{ or } \sin(\frac{x}{2})=-\frac{\sqrt{2}}{2}\]

Answer 31

how did you get a sqrt of 2 in the second part, wasnt there a 1 there before?

Answer 32

\[\sqrt{\frac{1}{2}}=\frac{\sqrt{1}}{\sqrt{2}}=\frac{1}{\sqrt{2}} =\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2} \\ \]

Answer 33

oh okay

Answer 34

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

Answer 35

Yea thats a good thing to keep a note of

Answer 36

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

Answer 37

\[\text{ we have two equations \to solve } \text{ Let } u=\frac{x}{2} \\ \sin(u)=\frac{\sqrt{2}}{2} \text{ or } \sin(u)=\frac{-\sqrt{2}}{2} \\ u=\frac{\pi}{4}+2n \pi , \frac{-\pi}{4}+ 2n \pi \text{ or } u=\frac{2 \pi}{3}+2 n \pi , \frac{5\pi}{4}+2 \pi n \\ \text{ or condensing this a little } \\ u=\frac{\pi}{4}+\pi n \text{ or } \frac{-\pi}{4} + n \pi \\ \text{ now remember we wanted } \frac{\pi}{4} \le x \le \frac{7 \pi}{4} \\ \text{ but we solved for } u \\ \\ \text{ recall } u=\frac{x}{2} \\ \text{ so } x=2u \\ \text{ so we need to find } u \text{ above such that } \\ \frac{\pi}{4} \le 2u \le \frac{ 7\pi}{4} \\ \text{ dividing 2 on all sides gives } \\ \frac{\pi}{8} \le u \le \frac{ 7\pi}{8} \]

so before we go further and solve for x
can you solve for u on the interval [pi/8,7pi/8]

Answer 38

wow, didnt think of splitting the equation into 2 separate parts. But yell ill solve for u on the interval

Answer 39

I dont know how to find where pi/8 and the other value is onthe unit circle

Answer 40

\[\text{ we are solving } \sin(u)=\frac{\sqrt{2}}{2} \text{ or } \sin(u)=\frac{-\sqrt{2}}{2} \text{ on } u \in [\frac{\pi}{8},\frac{7\pi}{8}] \\ u=\frac{\pi}{4}, \frac{3\pi}{4}\]

Answer 41

now since u=x/2
we need to solve:
\[\frac{x}{2}=\frac{\pi}{4} \text{ and also solve } \frac{x}{2}=\frac{3\pi}{4}\]

Answer 42

so x= 2pi/4 and 6pi/4

Answer 43

reducing
x=pi/2 or x=3pi/2

Answer 44

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

Answer 45

do i plug the x values in the derivative or the original function?

Answer 46

also for the above equation if it would have made things easy we could have also used the half angle identity for sin \[\sin^2(\frac{x}{2})=\frac{1}{2} \\ \frac{1}{2}(1-\cos(x))=\frac{1}{2} \\ \text{ multiply both sides by } 2 \\ 1-\cos(x)=1 \\ \text{ subtract 1 on both sides } -\cos(x)=0 \\ \cos(x)=0 \\ \text{ this probably would have been easier for you \to solve }\]

Answer 47

no you plug into the original function
the derivative will only tell you if the slope is positive,0, negative ,or non-existing

Answer 48

easier in general not just you lol

Answer 49

\[g(x)=x+\cot(\frac{x}{2}) \\ g(\frac{\pi}{4})=\frac{\pi}{4}+\cot(\frac{\pi}{8}) \\ g(\frac{\pi}{2})=\frac{\pi}{2}+\cot(\frac{\pi}{4}) \\ g(\frac{3\pi}{2})=\frac{3\pi}{2}+\cot(\frac{3\pi}{4}) \\ g(\frac{7\pi}{4})=\frac{7\pi}{4}+\cot(\frac{7 \pi}{8})\]

Answer 50

so you just might want to whip out your calculator and plug those into see which gives the highest and lowest output

Answer 51

okay, I see

Answer 52

This problem required a lot of work...I thought it wouldve been easy, I guess I have to review the concept behind this problem

Answer 53

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

Answer 54

half angle-identity *

Answer 55

but either way will suffice

Answer 56

got it, thanks!

Answer 57

cool stuff

Answer 58

peace

Answer 59

Step1- find critical values and replace them in original function
Step2- replace endpoints as well
Step3- check your highest and lowest values
Done :)