How do you find the domain just by looking at the equation?

1. Feb 23, 2005

gillgill

How do you find the domain just by looking at the equation?

ex1) x^2/ x^4+4x-5
2) t/√(t^2-5t+6)
3) √[2-√(4-x)]

2. Feb 23, 2005

arildno

What must you require of the argument of the (real) square root function?

3. Feb 23, 2005

dextercioby

Take the first.What's the point in which the division (employed by the fraction) has no meaning...?

Daniel.

4. Feb 23, 2005

gillgill

when x^4+4x-5=0

5. Feb 23, 2005

arildno

gillgill:
LEARN TO WRITE MATHS PROPERLY!! :grumpy:
For 1) Did you mean x^2/(x^4+4x-5), x^2/(x^4+4x)-5, x^2/x^4 +4x-5
Do you understand what I'm talking about?

6. Feb 23, 2005

HallsofIvy

Staff Emeritus
Assuming that your first function was x^2/ (x^4+4x-5) (do you see why I put in the parentheses?) then, yes, the denominator will be 0 when x^4+ 4x- 5= 0 and, since you cannot divide by 0, the domain is "all real numbers except solutions to x^4+ 4x- 5".

In problem 2, the quantity inside the square root cannot be negative or zero (since we are dividing by it).

In the last one, certainly we must have 4- x>= 0 (so x<= 4) in order to be able to do that square root but we must also have 2- sqrt(4-x)> 0. To determine what restriction that puts on x, look at 2- sqrt(4-x)= 0. That is the same as 2= sqrt(4-x) and, squaring, 4= 4-x which is the same as x= 0! If x< 0, then 4-x> 4,sqrt(4-x)> 2 and 2- sqrt(4-x)< 0. The domain is 0<= x<= 4.

7. Feb 23, 2005

dextercioby

It's not the one with the 5 in the denominator,i hope,else he would have to solve that quartic...Actually a cubic,because "+1" is a sollution...

Daniel.

8. Feb 23, 2005

arildno

In addition to HallsofIvy's suggestions, I would like to say that what they are asking you to find, is the MAXIMAL domain of the (real) functions over the real numbers.