Simplifying Precalculus Equations and Identifying Function Properties

[Caldus]

Need some clarification as to whether the following can be simplified any further (teacher takes off points for not being completely simplified) or how to simplify it further:

1. -3x + 5 + (x + 3)^1/2

2. -3x + 5 - (x + 3)^1/2

3. Domain of (3x + 5) / (x + 3)^1/2 <-- Is it x > -3?

4. Range of (-3x + 8)^1/2 <-- Is it y > 0?

5. ((-3x - 3h + 5) - (-3x + 5)) / h

6. Is the inverse of -3x + 5:
(x - 5) / -3?

7. Is this function even, odd, neither, or both? (I put neither):

v(x) = x^5/3

 

[StephenPrivitera]

1,2,4 and 6 look fine to me.
3) To find the domain of a rational function (one that is in the form of a fraction), set the denominator equal to 0. The denominator cannot be zero obviously, so this x-value is not in the domain. The domain is then the intersection of the domain of the numerator with the domain of the denominator minus the x-values for which the fraction is undefined (eg: if the domain of the numerator is x={1,2,3,4,5} and the domain of the denomator is x={1,2,5,8,9} but at x=2 the denominator is zero, then the domain of the function is x={1,5}).
5)Distribute the negative sign on the expression -(-3x+5).
7)I think this depennds on whether you mean y(x)=x^5/3 or (1/3)x⁵, but maybe not. Remember, to figure out if it is even or odd use the definitions:

f(-x)=f(x) <--> f(x) is even
eg: f(x)=x²
f(-x)=(-x)²=x²=f(x)

f(-x)=-f(x) <--> f(x) is odd
eg: f(x)=x³
just plug in -x for x and see what happens