Question

Graph y=(e^x)(sinx), find zeros, max/mins, and inflections points between or equal to -pi and +pi

Answer 1

to find the max/min, you basically have to equate the first derivative to zero. that will give you a value of "x". now after having done that, find the second derivative. if it is grater than zero, it will have min. value, if it is less than zero, it will have max. value

Answer 2

@1ace do you get it?

Answer 3

ok so i got y' = e^x(sinx+cosx) but i dont understand how to get max/ min points for that one

Answer 4

now equate that to zero.

Answer 5

e^x cannot be zero. so, sinx = - cosx
tanx = -1
x= tan^-1(-1)

Answer 6

!! oh wow i never saw that!
ok so x= 3pi/4 and 7pi/4 ?

Answer 7

only 3pi/4. 7pi/4 is out of the given limits

Answer 8

ooh ok
so for the second derivative y''= e^xsinx + e^xcosx - e^xsinx + e^xcosx ?
could i cancel out the sinx's there?

Answer 9

so for this one i would find cosx=0
and x= pi/2 and -pi/2 ?

Answer 10

so the given function has a min. value

Answer 11

no. the second derivative is greater than zero, so the function has a min. value

Answer 12

you don't equate the second derivative to zero.

Answer 13

but dont i equate y'' to zero to find inflection points?

Answer 14

fro the point of inflection, yes.

Answer 15

*for

Answer 16

right so y''= e^x(2cosx) = 0
where cosx = 0, then x= pi/2 and -pi/2 ?

Answer 17

yes

Answer 18

ok so in total i have 4 points max/min and inflection points i think i get this question now. Thanks !!

Answer 19

no problem