I need help with these problems for a test.

1. Find the inverse of the one-to-one function.
f(x)=5x+1/4.

2. The height of the water, H, in feet, at a boat dock t hours after 7 am is given by E=7+4.4cos pi/35 t, where t is time measured in seconds. Find the period.

3. Solve the equation on the interval [0, 2pi).
a) sin^2x - cos^2x = 0

b) sin4x=(sq3)/2

4. Factor completely, or state that the polynomial is prime.
27y^4-147y^2

1. Here is a neat way to find the inverse of a linear function.

perform the inverse of the operations in reverse order

original: we multiply by 5 then add 1/4
inverse: we subtract 1/4, then divide by 5
f^-1(x) = (x - 1/4)/5
= (4x-1)/20

check: pick any value of x, e.g. x = 8
f(8) = 40+1/4 = 161/4

f^-1(161/4) = (4(161/4) - 1))/20
= 160/20 = 8
checks out!

2. The period of E = 7+4.4cos(π/35)t

= 2π/(π/35)
= 70

3. sin^2x - cos^2x = 0

sin^2x - (1 - sin^2x) = 0
2sin^2x = 1
sin^2x = 1/2
sin x = ± 1/√2

x = 45°, 135°, 225° or 315°
or
x = π/4, 3π/4, 5π/4 or 7π/4

4. sin 4x = √3/2
4x = 60° or 4x = 120°
or
4x = π/3 or 4x = 2π/3

so x = π/12 or x = 2π/12 = π/6

but the period of sin4x is 90° or π/2
so by adding π/2 to each of the above answers we can get more answers until we go past 2π

π/12 + π/2 = 7π/12
7π/12 + π/2 = 13π/12
13π/12 + π/2 = 19π/12 adding one more π/2 would be beyond 2π

now do the same with π/6 to get 3 more answers from that branch

altogether there will be 8 different solutions.

lastly, .....

27y^4-147y^2
= 3y^2(9y^2 - 49)
= 3y^2(3y+7)(3y-7)

Thanks

Sure, I can help you with these problems. Let's go through each one step by step to find the solutions.

1. To find the inverse of a function, we need to swap the x and y variables and solve for y. Let's start with the given function f(x) = 5x + 1/4.

Step 1: Swap x and y variables:
x = 5y + 1/4

Step 2: Solve for y:
x - 1/4 = 5y
(x - 1/4)/5 = y

So, the inverse of the function is:
f^(-1)(x) = (x - 1/4)/5

2. The given equation is E = 7 + 4.4cos(π/35 t). We are asked to find the period of the function.

The period of a cosine function is given by the formula: T = 2π/b, where b is the coefficient of the variable inside the cosine function.

In this case, the coefficient of t is π/35. Therefore, the period T is:
T = 2π/(π/35)
= 2π * 35/π
= 70

So, the period of the given function is 70.

3. Let's solve each equation one by one:

a) sin^2x - cos^2x = 0

We can use the trigonometric identity sin^2x - cos^2x = sin^2x - (1 - sin^2x) = 2sin^2x - 1 = 0

Solving 2sin^2x - 1 = 0:
2sin^2x = 1
sin^2x = 1/2
sinx = ±√(1/2)

The values of sinx that satisfy this equation are x = π/4, 3π/4, 5π/4, and 7π/4 within the interval [0, 2π).

b) sin4x = √3/2

We can use the inverse trigonometric function arcsin to solve for x.

Taking arcsin on both sides of the equation:
4x = arcsin(√3/2)
x = arcsin(√3/2)/4

The solutions within the interval [0, 2π) are x = π/12 and 7π/12.

4. Let's factor the given polynomial: 27y^4 - 147y^2

We can factor out the common term 3y^2:
3y^2(9y^2 - 49)

Now, we have a difference of squares:
3y^2(3y + 7)(3y - 7)

So, the given polynomial is factored completely as 3y^2(3y + 7)(3y - 7).

I hope this helps! Let me know if you have any further questions.