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

[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)]

 

[arildno]

Take your last expression.
What must you require of the argument of the (real) square root function?

 

[dextercioby]

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

Daniel.

 

[gillgill]

when x^4+4x-5=0

 

[arildno]

gillgill:
LEARN TO WRITE MATHS PROPERLY!
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?

 

[HallsofIvy]

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.

 

[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.

 

[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.