# I face some difficulties to find the domain

1. Oct 11, 2009

### r-soy

I face some difficulties to find the domain

i try solve it

thanks

2. Oct 11, 2009

### Staff: Mentor

It can sometimes take several hours for attachments to be approved, so for quicker responses, it's best to type your questions out so that we can see them right away.

3. Oct 11, 2009

### r-soy

I think attachments are approved now >>

4. Oct 11, 2009

### Bohrok

The domain of problem 46 is not all real numbers. For √x, domain is x≥0. Use that for √(-1-s).

The domain of 47 is larger than just w≥4; perhaps you were still thinking about domains with square roots. Try thinking about the domain of 1/x, or where it is undefined, and then think about the domain of (2x+3)/(w-4) again.

I can't quite make out your answer for the domain of 51, but it doesn't look right. But let's look at what you have with (y + 1)(y - 3)≥0. You know where the parabola crosses the x-axis and that it opens upward since the leading coefficient is positive. What part(s) of the graph will be above the x-axis, or ≥0?

Last edited: Oct 11, 2009
5. Oct 11, 2009

### Anakin_k

As Bohrok said, if the domain was all real numbers for √(-1-s), it would be incorrect. Say s is equal to 2. -1-2 = -3. √-3 is an imaginary number. You want to make sure whatever is under the square root bracket is equal to 0 or more.

6. Oct 11, 2009

### oNothinGo

53. x^2-4x+5 and x^2+2x+3 can't be factor the way you did.

7. Oct 11, 2009

### Staff: Mentor

The domain for f(x) = √x is x >= 0.

8. Oct 11, 2009

### Bohrok

I hit the wrong two-key combination instead of Ctrl-V. Corrected.

9. Oct 12, 2009

### r-soy

46 )
1-s ≥ 0
-s ≥ - 1

47) i think domain all real numer

help me .>

10. Oct 12, 2009