Question

Help confused with applying cosine and sine functions

Answer 1

The number of hours of daylight in Boston is given by\[f(x)=3\sin(\frac{ 2\pi }{ 365 }(x-79))+12\], where x is the number of days after January 1. What is the most amount of daylight that Boston will experience?

Answer 2

With x does it mean 364?

Answer 3

Dude what do you mean ? by with x does it mean 364 ?

Answer 4

It says x is the number of days after january 1st

Answer 5

so there is 365 days in a year so it would be 364 then

Answer 6

remember that the "sine" function fluctuates between -1 and +1. so,
the extreme values for f(x) will be:
1) \(f(x)=3(1)+12\)
or
2) \(f(x)=3(-1)+12\)
which one gives you a bigger value?

Answer 7

The bigger value would be the first one

Answer 8

I was going to get the derivative and check where is the highest value for F(X)

Answer 9

because 1 is 15 and 2 is 9

Answer 10

you could do that and verify. Derivative method should give you a fool-proof solution. But, to me, this sounds pretty logical. Just verify both ways. Would be interesting.

Answer 11

always shoot for simplicity. do not complicate math. :)

Answer 12

Well Could you explain derivative method to me s: I have never heard of this

Answer 13

Sure:
1) the first derivative of a function gives the slope of tangent for that function.
2) at maxima or minima, the tangent is perfectly horizontal i.e., parallel to X-axis. Hence, its slope is "0"
3) therefore, if we find the derivative of a function and equate it to zero, we will get the conditions when the function is maximum or minimum.

Answer 14

either way, these points are called a critical points.
To check if a critical point is a maxima or a minima, we go for the second derivative.
"If the value of the second derivative at a critical point is
1. NEGATIVE => function has a maxima at that value of 'x'
2. POSITIVE => function has a minima at that value of 'x'
3. ZERO => the second derivative test fails to figure out what the critical point is. Most likely, it is neither and called a a "SADDLE" point

Answer 15

in the question, you are only asked for the maximum value of f(x) and not "when" you get a maximum value. So, I chose not to burden myself with derivatives.

Answer 16

DarkBlueChocobo, electrokid way of solving it was easier. Mine just will give you certain answer without guessing.

Answer 17

Could you show me an example problem for finding the maxima and minima Cause im still trying to grasp this

Answer 18

Also with derivatives I have not heard this term before so far in precal s:

Answer 19

ok. lets say you want to find when \(f(x)=x^3-x^2+1\) has a maximum value(s) and minimum value(s).
We do not know how many maxima or minima we have.

Answer 20

so,
step 1: Find the first derivative\[f'(x)=3x^2-2x\]

Answer 21

would the first derivative be 3x^2?

Answer 22

Nope.

Answer 23

step 2: equate it to zero and solve for 'x'
\[3x^2-2x=0\\x(3x-2)=0\\\fbox{x=0}\qquad\rm{or}\qquad3x-2=0\implies\boxed{x=\frac{2}{3}}
\]

Answer 24

Notice the following:
the degree of \(f(x)\) is 3 and it has "2" critical points. Always there will be \(n-1\) critical points for a function of order \(n\)

Answer 25

we good till here?

Answer 26

it would be 2/3 right?

Answer 27

no we have two critical points from step 2 above. we do not know at which value of "x" will \(f(x)\) be a maxiumum and at which a minimum. but for sure, at one value, we have a maximum and at other, a minimum. That is what the derivative test tells us.

Answer 28

we find in step 3.
kapeesh?

Answer 29

Alright S: ugh im just confused

Answer 30

which part?

Answer 31

I am confused with derivatives

Answer 32

which is the first and second

Answer 33

you good till step 1 and step 2?

Let me summarize the method:
step 1: we found the first derivative
step 2: we found critical points
step 3: we take another derivative,. Since this will be the second time we do it, it is called the second derivative.
step 4: check each critical point from step "2" to decide if we have a maxima or a minima at that point

Answer 34

so far, we have not found the second derivative yet. we good?

Answer 35

Well I mean what is a derivative is my question

Answer 36

hmm. did you take calculus I?

Answer 37

No I am in precal

Answer 38

ok. so, you are learning limits, right?

Answer 39

Well atm learning a bit of trig so Last unit i got introduced to all the functions

Answer 40

This unit is graphing said functions

Answer 41

and now applying them

Answer 42

Aha!
that is exactly why they asked you what the maximum value is but not "when" it is!

Answer 43

ohhh

Answer 44

s: so Is it good that i dont know this yet lool

Answer 45

yep

Answer 46

But what exactly is a derivative though

Answer 47

you were expected to know that a sine function or a cosine function can have a maximum of either +1 or -1 and use that so answer the question.

Answer 48

simply, a derivative tell you "how fast a function increases or decreases"!!

Answer 49

if a function remains the same (no change), the derivative is zero!

Answer 50

So you basically set sin to 1 or -1?

Answer 51

thats how you solved this? egh

Answer 52

yep