explain how to find the maximum value for each function and determine which function has the largest maximum y-value.

- anonymous

- schrodinger

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- anonymous

##### 1 Attachment

- anonymous

- UnkleRhaukus

The function is \[g(x) = 2\cos(2x-\pi)+4\]
This will have maximum values when the cosine term is at maximums

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## More answers

- anonymous

f(x) = −4(x − 6)^2 + 3

- UnkleRhaukus

do you know what the largest value of \(\cos\theta\), can ever be? (for all possible angles \(\theta\) )

- anonymous

no

- UnkleRhaukus

remember that cosine is the ratio of the adjacent side in a right angled triangle, to its hypotenuse,
(note that the hypotenuse is always the largest side)

- anonymous

ok ok , i get that

- UnkleRhaukus

|dw:1440303568385:dw|

- UnkleRhaukus

The cosine function (like the sine function ) oscillates between its maximum and minimum values ; ±1

- anonymous

okk

- anonymous

So the maximun for g(x) would be 1?

- UnkleRhaukus

not quite,
that max for cos theta,
and hence the max for cos(2x-π) is 1

- UnkleRhaukus

so the max for g(x) = 2cos(2x-π) + 4
is gmax = 2(+1) + 4
=

- anonymous

so 7 lol

- UnkleRhaukus

(whereas the minimum is
gmin = 2(-1) + 4
= -2+4
= 2
)

- UnkleRhaukus

wait
gmax = 2(+1) + 4
= 2 times 1 + 4
=

- anonymous

oh i was thinking adding. silly me lol. so 6

- UnkleRhaukus

notice that we can see in the plot, that the blue line is oscillating between 2 and 6, which agrees

- UnkleRhaukus

so now lets look at \[f(x) = −4(x − 6)^2 + 3 \]

- anonymous

how would we set that up

- UnkleRhaukus

What can you tell me about f(x)?

- UnkleRhaukus

(what is the parent function of f )

- anonymous

i dont get it?

- UnkleRhaukus

its a second order polynomial, right?
so its a quadratic equation, and will look like some sort of parabola

- anonymous

is this calculus? if so, take the derivative of the function, and set to 0 to find where it has a max (or min).
However, you do not need calculus. The first function,
f(x)=-4(x-6)^2 + 3
is a parabola ( in the shape of a "frown" ) with a max value at its vertex.
because they gave you the equation in "vertex form"
y = a(x-h)^2 + k , you can read off the vertex to be (h,k)
In this case, the vertex is at (6,3)
for the second function,
g(x)=2cos(2x-pi)+4
you should know that the max value of the cosine is 1, so the max will be
g(x)= 2*1 + 4 = 6

- UnkleRhaukus

@TavTav are you still here?

- anonymous

yes sorry

- anonymous

so i got g(x) but how would you get f(x)

- UnkleRhaukus

given that f(x) is some sort of parabola, do you expect f to have both max and min values?

- anonymous

no

- UnkleRhaukus

\[f(x)=-4(x-6)^2 + 3\]
which is a bit like: \(-x^2\)
which do you expect: a max, or a min value?

- anonymous

max

- UnkleRhaukus

good, and the max will be when the terms under the square, will be what value?

- anonymous

the max value would be the 3?

- UnkleRhaukus

yeah, the max is when the terms under the square is zero,
\[f_\text{max}=-4(0)^2 + 3\\
\qquad=-0+3\\
\qquad=3\]

- anonymous

thank you.
i need to have it in complete sentences so can you help me form those?

- anonymous

for both g(x) and f(x)

- UnkleRhaukus

show me what you got

- anonymous

uum ok

- anonymous

lol

- UnkleRhaukus

The general idea was that we considered the parent functions
\(G(X) = \cos (X)\), and \(F(X)=-X^2\), respectively

- anonymous

I know how we got g(x) i just dont know how to put it in words
for f(x) we used the zeros of the equation to find the maximun value?

- UnkleRhaukus

we didn't really use the "zeroes of the equations", they are different things

- anonymous

ohh , uhh.
so how would you word it

- anonymous

that we compared the functions to their parent functions to get the maximum value

- UnkleRhaukus

yeah, something like that,
we know how to find the max/min of the parent functions, and then we apply these results to our particular functions

- UnkleRhaukus

scaling and shifting as specified

- anonymous

so i can use that as the answer?

- UnkleRhaukus

kind of. you might like to write separate paragraphs for each function

- anonymous

okok , thank you so much!! you made this easy to understand.your a lifesaver lol

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