How do you find the maximum value for each function?

Question

tkhunny · Answer

Do we get to use the 1st Derivative?

anonymous · Answer

What? Lol, sorry I'm not too good at math.

anonymous · Answer

Is there like a formula for that? or

anonymous · Answer

do you know limits?

anonymous · Answer

um, no.. This is trig I believe..

tkhunny · Answer

Fair enough.  You should have studied parabolas in Algebra 2.  Did you?  Find the vertex of the Downward Opening parabola and you will be done with that one.  It is in Vertex form.

tkhunny · Answer

What is the maximum value of the cosine?

anonymous · Answer

2?.. i don't know.

anonymous · Answer

How would I find that

tkhunny · Answer

Did you find the vertex?  It's kind of staring at you in this vertex form.

$-1\le\cos(x)\le1$

What is the maximum value of the cosine?

tkhunny · Answer

f(x) maximum (6,3) -- Perfect.

g(x) Did you read my hint?  Definitely not 4pi.

anonymous · Answer

Would it be 1?

tkhunny · Answer

Yes, it would be.

For what value of x does the cosine function take on the value 1?

anonymous · Answer

I'm not sure what you mean.. Would I look at the unit circle or no.. /;

tkhunny · Answer

Yes.  Reference to the Unit Circle will answer my question.  Where is the cosine +1?

anonymous · Answer

(1,0)?

anonymous · Answer

I just graphed it on geogebra and the y cordinate is 6 but I'm not sure what the x cordinate would be.

dumbcow · Answer

yes max value is y=6 because cosine has amplitude of 2 and is shifted up 4  (4+2=6)

cos(2x-pi) = 1   (this makes y=6)
cosine is 1 when angle is 0 or multiple of 2pi

2x-pi = 0
x = pi/2 + 2pi*n

dumbcow · Answer

no @tkhunny helped you correctly for first one ... vertex of parabola is (6,3) so max is y=3

dumbcow · Answer

yes you can tag me and if im available i will try to help

anonymous · Answer

Okay thank you :D

tkhunny · Answer

$2x-\pi = 0+2k\pi \longrightarrow 2x = \pi + 2k\pi \longrightarrow x = \dfrac{\pi}{2} + k\pi$

Not quite what dumbcow produced.