Question

How would I get the graph for this and what would it look like?
g(x)=
e^((−3x+2sinx)/6)
------------------
(3+2cosx)

Answer 1

What, again.
Simplest is to feed it to Wolfram.
Else note that it always positive, calculate maxima/minima, end behaviours etc

Answer 2

I have done that but I need to understand how i get to the answer through calculation. Could you guide me through it.

Answer 3

I am a bit lost on this problem

Answer 4

You can see that the top and bottom are positive so you need not concern yourself with anything below the x-axis.

Answer 5

I forget what they were but you get the derivatives and possible min/max (then test either side for increasing/decreasing or else use second derivative to check if min or max.

Answer 6

Now, what happens if x= 0?

Answer 7

Now consider what happens for increasing negative and positive vales of x.

Answer 8

Sketch it.

Answer 9

Do a couple more values if necessary.

Answer 10

Oh and I forgot, there is a value of x that makes the denominator zero, need to look at that as well.

Answer 11

I get to this point but can't make sense of what they mean in terms of the graph. (sinx -5/2) and (sinx-1/2). These are meant to be the stationary points. But I can visualise it.

Answer 12

I don't understand what you mean? How can sinx -1/2 be a stationary point?

Answer 13

I mean stationary points are when x is 1/2 and x is 5/2.

Answer 14

OK, I assume that those are correct, now we need to know whether they are max or min (you can test either side for positive/negative or use second derivative)

Answer 15

You then need to calculate y at the min/max to be able to plot those points.

Answer 16

Do they seem right to you? I will now test those point.

Answer 17

Wait a minute, I will get a graph from Wolfram to check.....

Answer 18

thanks

Answer 19

http://www.wolframalpha.com/input/?i=plot+ [g%28x%29%3D++e^%28%28%E2%88%923x%2B2sinx%29%2F6%29+%2F%283%2B2cosx%29%2C{x%2C-10%2C5}%2C{y%2C-1%2C10}

Um, can you check if I have the correct equation, I am seeing max/min somewhere else

Answer 20

Paste
plot [g(x)= e^((−3x+2sinx)/6) /(3+2cosx),{x,-10,5},{y,-1,10}

into Wolfram box

Answer 21

plot [g(x)= e^((−3x+2sinx)/6) /(3+2cosx),{x,-10,5},{y,-1,10}]

Answer 22

Here it is

Answer 23

OK, I found it now..

Answer 24

Ok, the equation is correct at Wolfram.

So where did you get 1/2 and 5/2 from?

Answer 25

I got it from working out 4sinx^2 +12sinx -5 =0 I used factoring and quadratic formula to get the stationary points

Answer 26

4sinx^2 +12sinx + 5 =0 not -5

Answer 27

And I don't get 1/2 and 5/2

Answer 28

is it not -4sinx^2 -12x +5 or 4sinx^2 +12sinx -5. Are they not the same?

Answer 29

Above, you said "4sinx^2 +12sinx -5 =0" it is +5 not -5
It also says 4sinx^2 +12sinx -5 =0 in the doc you sent me.

Answer 30

It also says 4sinx^2 +12sinx +5 =0 in the doc you sent me.

Answer 31

Anyway, this is a periodic function

Answer 32

there is a minus sign sign just after g'(x). What happens to that?

Answer 33

Maybe that is why I am getting a bit lost. Because the exercise up to now have not given me any exercise in this type of problem.

Answer 34

If you want to you can absorb it into the eqaution to get -4sin^2 x +12 sin x -5

Answer 35

Anyway, here is Wolfram
plot[4sinx^2 +12sinx +5 =0 ,{x,-10,10},{y,-10,10}]

Answer 36

can I just clarify something

Answer 37

Now you have me doing it!
plot[4sinx^2 -12sinx +5 =0, {x,-10,10},{y,-10,10}] is correct.

Answer 38

I'm sorry, I really do not want to do any more of these computations.

Answer 39

\[- ((4sinx-12sinx +5)\exp((-3x+2sinx)/6)/(6(3+2cosx)^2)\]

Answer 40

−((4sinx−12sinx+5)

Answer 41

It is sin^2 x not sin x

Answer 42

Is this what you have in the file I sent you for the denominator?

Answer 43

(4 sin^2 x -12 sin x +5) and there is a minus sign in front of the whole derivative which you can absorb anywhere you want to (or not).

Answer 44

Anyway, we are spending too much time and getting nowhere, so I have to go now, i am sorry.

Answer 45

Ok. Thanks for your help

Answer 46

OK, I hope you can solve it (it really is not too difficult once you concentrate on it)