Question

How to find the maximum value of g(x) = 2 cos(2x - π) + 4?

Answer 1

what is the maximum value of cos?

Answer 2

isn't one?

Answer 3

so the max value is ¨6

Answer 4

@RaphaelFilgueiras can you explain more, how to know that it's 6? :o
I forgot this.

Answer 5

-1<=cos(u)<=+1, then the max value of g(x), is reached when cos(2x-pi)=1

Answer 6

ohh thanks xD @RaphaelFilgueiras

Answer 7

I was thinking in terms of f(x) = a (cosb(x-h)) + k
max= a+k

Answer 8

you are welcome

Answer 9

you can differentiate it too

Answer 10

but idk if it's always true,
thanks for proving it thought! <3

Answer 11

wait so how do y'all know it's 6? @study100 @RaphaelFilgueiras

Answer 12

tried googling it and it said if a < 0 then the max is k, but it didn't explain how to find the max value when a > 0

Answer 13

http://web.psjaisd.us/auston.cron/ABCronPortal/GeoGebraMenu/GeogebraFiles/trigonometry/graphics/cosineDefn.PNG
refer to image, you see that a+k = max point
see the green and purple line add up to max

Answer 14

a= 2
k= 4 so
2+4 = 6
but raphael has a really good way to prove that too.

Answer 15

so this method would only be used if a > 0? @study100

Answer 16

This is the other equation I needed to find the max value of. f(x) = -4(x - 6)^2 + 3 .

I said it was 3, is that correct? Or would it be -1?

Answer 17

oh when a is less than 0 all you need to do is forget about the negative sign in a.
maximum of ALL COS FUNCTION = absolute value of a plus c = l a I + k

Answer 18

I assume you know absolute value?

Answer 19

plus k***

Answer 20

Proof, here I graphed -acos... for you

Answer 21

still 6, see? :DD

Answer 22

y = a(x-h) ^2 + k
max is vertex
vertex = (h,k)
h= 6
k= 3

So the exact point is at (6,3) for max

Answer 23

Ahhh I'm so confused lol

Answer 24

Look at 1 on this page http://www.wikihow.com/Find-the-Maximum-or-Minimum-Value-of-a-Quadratic-Function-Easily

Answer 25

first or second?

Answer 26

That's why I thought when a < 0, the max value is automatically k

Answer 27

in the second one the first therm is always negative then the max is when it's zero, so the max is 3

Answer 28

Method 2

Answer 29

So for this equation the max value WOULD be 3, right? f(x) = -4(x - 6)^2 + 3

Answer 30

yes

Answer 31

And so the max value of g(x) = 2 cos(2x - π) + 4? is 12?

Answer 32

if you use the y = c- (b^2/2a)
it will generate the same answer