Question

Find the critical numbers of the function. (Enter your answers as a comma-separated list. Use n to denote any arbitrary integer values. If an answer does not exist, enter DNE.) g(θ) = 36θ − 9 tan θ

Answer 1

hey @FATIMA955 so any idea where to start?

Answer 2

Do you know how to get critical numbers and values?

Answer 3

??

Answer 4

well find the derivative of \(g(\theta)\)

Answer 5

do it now!! hehe

Answer 6

Here something that you should know by now
to find crictical point you need to find the derivative
then set that derivative equal to zero and solve

Answer 7

Did you learn that
in the lecture?

Answer 8

f is continuous on the interval [a,b] and differentiabale on (a,b)
we say that c is a critical number if and only if f'(c)=0 or f'(c) does not exist

Answer 9

come on! do something please lol
I can't help you if you are just watching the screen there

Answer 10

hey @FATIMA955, by now, you should be able to do any derivatives

Answer 11

hey yeah sorry

Answer 12

how am i suppose to know all derivates havent studied math for past 3 years yeah im done with the derivative part though looking at the tips u have given

Answer 13

i need to find the value for theta thts it right?

Answer 14

oh ok, but if you are taking calc now you should review all those stuff again
so \(\huge \rm g'(\theta)=-9sec^2(\theta) \)
set it equal zero \(\huge \rm g'(\theta)=0 \Longrightarrow -9sec^2(\theta)=0\)

Answer 15

yes you need to solve for theta

Answer 16

you need some precalc here! trigonometry

Answer 17

so if we have \(\huge \rm -9sec^2(\theta)=0\) then \(\huge \rm sec(\theta)=0\)
now you should ask when does sec give zero

Answer 18

you might need to relook how the graph of sec looks like

Answer 19

this cant exist right

Answer 20

http://www.calculatorsoup.com/images/trig_plots/graph_sec_pi.gif

Answer 21

yes! secant can never be zero

Answer 22

since cos x =1/sec x having it =0 would mean that the cosine is undefined for some value .so we conclude the ans doesnt exist??

Answer 23

So what do you think

Answer 24

hmm no!
what we do is consider the values where g' doesn't exist
meaning the values that make secant undefined
it is not cosine who is undefined for some values but secant
cosine is perfectly defined, you need to think what values make 1/cosx not defined

Answer 25

after finding the dvt we get [36-9\sec^2\]

Answer 26

recall that f'(c) does not exist implies that c is critical point
when you have f'(c)=0 c could be either maximum or minimum

Answer 27

no? i give you the derivative already

Answer 28

yeah u did but it was 36 theta-9tan theta

Answer 29

No it was \(g'(\theta)=-9sec^2\theta\)

Answer 30

Are focusing here or not! i wrote it above
that part is done! now we are looking for the points that make g' does not exist

Answer 31

the first thing i did is find \(\huge \rm g'(\theta)\)
then i set up \(\huge \rm g'(\theta)=0\)

Answer 32

now we are in the final step to get the values of theta that make \(\huge \rm g'(\theta) ~undefined \)

Answer 33

yes yes i get ittt..

Answer 34

if you look back here in the graph you should be able to tell
what values make secant undefined which make g' also undefined

Answer 35

http://www.calculatorsoup.com/images/trig_plots/graph_sec_pi.gif

Answer 36

to make it easy for you the values are \(\huge \rm ...-5\pi/2-3\pi/2,-\pi/2,\pi/2,3\pi/2,5\pi/2...\)

Answer 37

the values continue from left and to right as well
all of those values make secant undefined

Answer 38

ohh ok thanksss..

Answer 39

well you are not finished yet lol
and np!

Answer 40

hahah yess i know ..

Answer 41

ok! good then

Answer 42

good luck i have to go

Answer 43

oh ok sure thankss alot...

Answer 44

^_^