There are many examples of graphs that have period 60 and whose y-values vary between -8 and 8. First, find a formula for such a function f, given that f(0)=8. Next, find a formula for such a function g, given that g(0)=0.

For your f, also find the three smallest positive x that makes f(x)=5

I don't know where to start for this problem...

for functions like y = a sin(kx) and y = a cos(kx)

the amplitude is a, that is, y lies between -a and +a
and the period is 2π/k

so for yours 2π/k = 60
k = 2π/60 = π/30
we are also supposed to have f(0) = 8, so it could be a cosine function
f(x) = 8cos(πx/30)

for g(0)=0, let's use a sine curve, since it starts at (0,0)
g(x) = 8sin(πx/30)

for f(x) = 5
8cos (πx/30) = 5
cos (πx/30) = .625
set your calculator to radians (RAD),
and I get πx/30 = .89566... , stored it in memory
x = (30/π)(.89566...) = 8.553 correct to 3 decimals

but the cosine is positive in I or IV
so x = 2π - 8.553 = -2.2398.. , a negative, we don't want that
but if we add a period to it, we should get another
x = -2.2398 + 60 = 57.730
another solution would occur one period later, so
x = 60 + 8.553 = 68.553

so the three smallest are: x = 8.553, 57.730, and 68.553

Proof:
http://www.wolframalpha.com/input/?i=plot+f(x)+%3D+8cos(%CF%80x%2F30)+,+0+%3C+x+%3C+100

Okay, that makes so much more sense. Thank you so much!

Well, let's break it down!

For a function with a period of 60 and y-values varying between -8 and 8, we can start by considering the basic cosine function, which has a period of 2π. To stretch the period from 2π to 60, we can use the formula f(x) = A*cos(Bx), where A is the amplitude and B is the frequency.

To ensure that f(0) = 8, we can set A = 8. Now, the largest maximum value for f(x) is 8 and the largest minimum value is -8. Since the cosine function oscillates between -1 and 1, we need to scale our frequency to ensure the amplitude is multiplied by 8.

The frequency, B, can be calculated as B = 2π/60. So, our function f(x) would be f(x) = 8*cos((2π/60)*x).

For g(x), the same principles apply, but g(0) is given as 0. To achieve this, we can use the sine function instead of cosine. So, our formula for g(x) would be g(x) = 8*sin((2π/60)*x).

To find the three smallest positive x values for which f(x) = 5, we can set the equation 5 = 8*cos((2π/60)*x) and solve for x. However, since you asked for humor, here's a joke instead:

Why did the graph's mom get upset with it?

Because it always kept plotting against her!

To find a formula for a function that has a period of 60 and whose y-values vary between -8 and 8, we can start by considering a sine or cosine function. Both sine and cosine have a period of 2pi, which is equivalent to 360 degrees. To fit a period of 60, we can scale the function by dividing the angle by (2pi/60) or (360/60), which simplifies to pi/30 or 6 degrees.

Let's start with the function f(x) that satisfies f(0) = 8. We can use a cosine function for this case:
f(x) = A * cos(Bx) + C

Given that f(0) = 8, we have:
f(0) = A * cos(B * 0) + C = A * cos(0) + C = A + C = 8

Next, we know that the y-values vary between -8 and 8, so the amplitude (A) should be 8. This gives us:
8 + C = 8
Therefore, C = 0.

So far, we have f(x) = 8 * cos(Bx) + 0, which simplifies to:
f(x) = 8 * cos(Bx)

To find the value of B, we know that the period of the function is 60. The period of a cosine function is given by 2pi/B, so we have:
2pi/B = 60
B = 2pi/60
B = pi/30

Thus, the function f(x) that satisfies f(0) = 8 is:
f(x) = 8 * cos((pi/30)x)

Now, for the function g(x) that satisfies g(0) = 0, we can use a sine function:
g(x) = A * sin(Bx) + C

Given that g(0) = 0, we have:
g(0) = A * sin(B * 0) + C = A * sin(0) + C = C = 0

So, C = 0.

Now, to find the value of A, we know that the y-values vary between -8 and 8, so the amplitude (A) should be 8. This gives us:
8 * sin(Bx) = 8
sin(Bx) = 1

The smallest positive value of x that makes sin(Bx) = 1 is when Bx = pi/2. Simplifying, we have:
Bx = pi/2
x = (pi/2) / B
Since we know B = pi/30, we substitute this value in:
x = (pi/2) / (pi/30)
x = 15

Therefore, the three smallest positive x that make f(x) = 5 are: 15, 45, and 75.

To find a formula for a function with a period of 60 and y-values ranging from -8 to 8, we can use a trigonometric function such as sine or cosine. Since we want the function to have maximum value at x=0, we can use the cosine function.

Let's start with the function f(x). We know that f(0) = 8, so the amplitude of the function is 8. The formula for a cosine function with a period of 60 and an amplitude of 8 is:

f(x) = 8 * cos((2π/60)x)

Now let's find a formula for another function g(x) given that g(0) = 0. Again, we'll use a cosine function with a period of 60. Since g(0) = 0, the formula becomes:

g(x) = 8 * cos((2π/60)x - π/2)

This formula is shifted π/2 (90 degrees) to the right from the formula of f(x).

Now let's find the three smallest positive x values that make f(x) = 5.

To find the x values, we need to solve the equation f(x) = 5. Using the formula for f(x), we have:

8 * cos((2π/60)x) = 5

Divide both sides by 8:

cos((2π/60)x) = 5/8

Taking the inverse cosine (arccos) of both sides:

(2π/60)x = arccos(5/8)

Solving for x:

x = (arccos(5/8) * 60)/(2π)

Using a calculator or a computer, we can determine the value of arccos(5/8) is approximately 0.760.

Therefore, the three smallest positive x values that make f(x) = 5 are:

x1 = (0.760 * 60)/(2π) ≈ 6.093,
x2 = (0.760 * 60 + 60)/(2π) ≈ 18.093,
x3 = (0.760 * 60 + 120)/(2π) ≈ 30.093.

So, the three smallest positive x values that make f(x) = 5 are approximately 6.093, 18.093, and 30.093.