I need help on these questions..... notice on how I am very careful on my grouping so you don't get confused on the order of operations.....

1) If f(y) = (2-y) / (2+y), find [f(y + k) - f(y)] / k

This is my answer...... [(2-y-k) / (2k+yk+k^2)] + [(-2+y) / (2k+yk)]

The problem is that my answer seems wrong......the equation seems too long to be correct....can it be simplified even more or is my answer correct?

2) Simplify by factoring

2(x+4) [(2x-1)^ -3] + 6[(x+4)^ -2](2x-1)^2

I could have solved this one if it wasn't for the negative exponents...

3) Also, can someone explain this question for me......

Which (if any) of the following equations are valid for all theta?

a) sin(theta + 2pi) = sin theta

b)cos(- theta) = cos theta

c) sin(pi + theta) = - sin theta

All of those are correct by the way, but I just don't understand what the question is asking. It is asking which of the equations are true?

## Answers (3)

Question1

f(y+k) = (2-(y+k)) / (2+(y+k)) = (2-y-k) / (2+y+k)

f(y) = (2-y) / (2+y)

f(y+k) - f(y)

= [ (2-y-k)(2+y) - (2-y)(2+y+k) ] / (2+y+k)(2+y)

= [ -4k / ( ky+2k+y²+4y+4 ) ]

= [ -4k / (y+2)(k+y+2) ]

Now divide by k

= [ -4 / (y+2)(k+y+2) ]

Question2

Factored it is:

2 (x+4) (2x-1)² [ (2x -1)^(-5) + 3(x+4)‾³ ]

Question3:

a) True, because sin has period of 2π

b) True, because cos is an even function

c) True, because sin has different sign but same value in the (diagonally) opposite quadrant.

1) The answer I got was -4, by multiplying the top half of the equation by the LCD. When you cancel terms you end up with -4k/k, which is -4.

2) The answer I got was 2(x+4)(2x-1)^2[(2x-1)^-5+3(x+4)^-3]

Who cares about negative exponents? It wants you to factor.

3) a) Theta+2pi = Theta, so sin(theta)=sin(theta)

b) even/ odd properties

c) When you add pi to something, it goes to the opposite side of the unit circle, where the sin value will be the same, but it will be negative of where you started.

f(y) = (2-y) / (2+y)

f ( y ) = ( 2 - ( 2 + y ) + 2 ) / ( 2 + y )

f ( y ) = ( 4 - ( 2 + y ) ) / ( 2 + y )

f ( y ) = 4/( 2 + y ) - 1

f ( y + k ) = 4/ ( 2 + y + k ) - 1

now

f ( y + k ) - f ( y ) = 4/( 2 + y ) - 1 - ( 4/ ( 2 + y + k ) - 1 )

f ( y + k ) - f ( y ) = 4/( 2 + y ) - 4/ ( 2 + y + k )

put 2 + y = t

f ( y + k ) - f ( y ) = 4/t - 4/( t + k )

..........................= 4 ( t + k - t ) / ( t ( t + k ) )

..........................= 4 k / ( t ( t + k ) )

<=>

[ f ( y + k ) - f ( y ) ]/k = 4/ ( t ( t + k ) )

................................= 4/( 2 + y ) ( 2 + y + k )