Question

Find the locations and values of any global extrema of function f(x)=sin((pi/2)x) on each of the four given intervals.

a) [-2,2]
b) (-2,2)
c) [-1,1)
d) [0,+infinity)

Answer 1

In past problems, what have you done to find "global extrema" of a function on a given interval?

Answer 2

i found the first derivative and then the critical points for it

Answer 3

as the first step

Answer 4

so the first derivative would be f'(x)=cos(pi/2)((1/2)x)

Answer 5

i don't know about the critical points though :(

Answer 6

i don't know how to solve for them really with any trig formulas involved

Answer 7

double check your derivative..

Answer 8

so the derivative is actually cos((pi/2)x)((1/2)pi)

Answer 9

chain rule states that d/dx f(g(x)) = f'(g(x))*g'(x) right? what's the derivative of the inner function in this case?

Answer 10

I agree with LifeEngineer: While some of your work towards finding the derivative of f(x) is correct, some is not. Try again, show us what you've done, and then we'll give you further feedback.

Answer 11

Ok so I changed it where it is (1/2)*pi not (1/2)*x

Answer 12

precisely.

now as to finding the critical points, you have to set this equal to 0

just get rid of the pi/2 since its just a constant
so:
cos(pi/2x)=0

now the best way to think about this is to imagine the unit circle.
cosine of any angle is just the x position as you go around the unit circle


so if you want the cosine of an angle to be zero, what angle do you need?

Answer 13

think of what angles (in radians) would give you x = 0

Answer 14

You want the angle to be pi/2 or 3pi/2 since x=0 for each of those

Answer 15

exactly!

now realize that your function is cos(pi/2*x) not just cos(x)

so if you want the angle to be pi/2 or 3pi/2, then you need to solve for what x would actually be

Answer 16

i don't really know how to do that :(

Answer 17

or at least with trig functions i don't

Answer 18

okay, lemme see if i can restate that question...

you know cosine of pi/2 = 0 and cosine of 3pi/2 = 0. and looking furthur into the other parts of the problem, you might realize that any odd multiple of pi/2 will also give you a zero (eg pi/2, 3pi/2, 5pi/2... as well as negatives: -pi/2, -3pi/2)

so the function you are trying to solve is cos(pi/2*x)=0

to solve youd have to take the inverse cosine (AKA arccos) of each side. This is just a mathy way of asking yourself, "the cosine of what angle equals this"

in the case of cos(pi/2x)=0, taking the arccos of each side gives you
pi/2x=arccos(0)

now weve already said that to get cosine of an angle to equal 0, the angle has to be an odd multiple of pi/2
so pi/2x=n*pi/2 (where n is any odd scalar)

then you just have to solve for x and all the nasty trig is gone! divide both sides by pi/2 and realize that that made your numbers nice and easy.

x=n so x is any odd number (... -3, -1, 1,3,5...)

Answer 19

so the critical points would be n(pi/2) where n=x and x is any odd scalar

Answer 20

yep! and if you think about it, your original sine function oscillates infinitely between -1 and 1 right? so it makes sense that you have an infinite number of critical points

Answer 21

going back to the original equation, apply this finding to the bounds given. if you want to see which are maximums and which are minimums, try plugging a few values in and see if you can see a pattern! good luck :)

Answer 22

right :)

Answer 23

now i need to find the second derivative and check the critical points of the first derivative into the second one for values

Answer 24

so the second derivative is -sin((pi/2)x)((1/4)pi^2)