math pre cal help?

Question

1. find the zeros of y= 2x^3 - x

2. find th range of y= 3^x

3. find the domain of y= (x^2 +1)/ (x^2 - x - 6)

4. [ -9.998 ] = ?

Answer 1

1. find the zeros of y= 2x^3 - x = x( 2x^2 -1) =0
then x =0 or x = sqrt(1/2) or x = -sqrt(1/2)


2. find th range of y= 3^x
range y >0

3. find the domain of y= (x^2 +1)/ (x^2 - x - 6)
domain all x except when x^2 - x - 6 =0
x^2 - x - 6 = ( x-3)( x+2) =0 => x=3 or x = -2


4. [ -9.998 ] = -10

Answer 2

1) Set 2x^3 - x = 0 and solve for possible values of x. I'll give you that x=0 is a solution for free.

2) What possible values could y have? Can y be negative? Can y = 0?

3) Are there any values that you cannot use for x? (hint: Set the bottom equal to 0 and solve)

4) I'm not certain what you mean by the square brackets....