Question

Given the function defined by y = e^sin(x) for all x such that - pi< x <2pi
a. Find the x- and y-coordinates of all maximum and minimum points on the
given interval. Justify your answers.
b. On the axes provided, sketch the graph of the function.
c. Write an equation for the axis of symmetry of the graph

Answer 1

where are you stuck?

Answer 2

how do i write the equation for the axis of symmetry?

Answer 3

So, you can do a, b?

Answer 4

show me your graph, that is part b

Answer 5

If you don't know, just speak out. That's perfectly ok. That's why we are here to study.

Answer 6

I actually need help with finding the max and mins

Answer 7

ok, the amplitude of the function is the number in the front of sin, that is e. So that , max of function is e, min is -e
Note: e is a number = 2.718

Answer 8

i found the derivative and then found the zeros and plugged the zeros into the original equation to find the max and min

Answer 9

why? we have it from precalculus course. the amplitude of the trig function can be found on that way.
However, if you want to go around, it gives you the same answer, just waste your time.

Answer 10

ok, now, do it once;
y' = (esinx)' = e cos x = 0 iff x = pi/2 or x = 3pi/2

Answer 11

when x = pi/2 , sin x =1 then y = e
when x = 3pi/2 sin x = -1 , then y = -e

Answer 12

the same.

Answer 13

oh thats makes since. thank you

Answer 14

ok, you can do the rest, right? let me know if you need more help :)

Answer 15

okay so i have my graph but how do i find the axis of symmetry?

Answer 16

ok, I give you the trick: sin is the function whose symmetric is y-axis
cos is the function whose symmetric is x-axis
so that your problem has y-axis is the axis of symmetry and its equation is x =0

Answer 17

so i need an equation the reflect the y axis?

Answer 18

yes, that is x =0

Answer 19

so i could use e^sinx -1?

Answer 20

nope, the symmetry equation is totally separated from the original function. You get it after drawing the graph out. That is why it is part c after part b.

Answer 21

oh okay

Answer 22

could pi/2 work?

Answer 23

OH,,,, let me consider, since your function is restricted from (-pi/2, 2pi), hence the middle point is 5pi/4
so that the symmetric equation is x = 5pi/4
I am so sorry. I forgot the restriction.

Answer 24

i think the domain is actually -pi to 2 pi

Answer 25

damn!! my bad again. Ok, I apology!!
3pi/2

Answer 26

its not pi/2??

Answer 27

nope, let draw it out

@dan815 rescue me, please, I am carelessly working on this problem and carelessly gave out the wrong answer many times. Shame on me. After seriously consider it, part c gives me no symmetric equation on that interval. Can it happen?