Precalculus Help

  • Thread starter Caldus
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Main Question or Discussion Point

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
 

Answers and Replies

  • #2
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)=x5/3 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
 
Last edited:

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