Help with precalculus? really confused...?

Question

how do you find the domain of this function? or any function, for that matter.

f(x)= (x-1)/(x^2-1)

i'm just soo confused...

Answer 1

The domain for any function is the values of x for which the function is defined. for something simple, like f(x)=x, the domain is (-infinity, +infinity) because you can plug in any number for x, and find a y value. However, if the function was f(x)=1/x, then the domain would be (-infinity,0) and (0, +infinity), because you cannot divide by zero. The function is not defined over that value.

Answer 2

You find what makes the denominator a zero. The domain is then everything else.

Your denominator is (X^2-1). In other words it is X squared minus 1. So something squared minus one = zero is what we are looking for.

You can just plug in some numbers, try 2. 2 squared minus 1 is 3-not what we want.

1 squared minus 1 is 0 and -1 squared minus 1 is 0. Bingo!

You can also solve the denominator for X by setting it equal to zero-> X^2-1=0 -> X^2+1-1=0+1 -> X^2=1 -> so X=-1 and X=1 Bingo!

The domain is then all real numbers except 1 and -1

I can't explain things good on the computer but I hope this helps.

Answer 3

the domain is the possible numbers that x is allowed to be. for example, in this function we have a fraction. and we know that in fractions the denominator is not allowed to equal 0. so...

x^2 -1 cant = 0

solve for x by factoring:

(x+1)(x-1) cant=0

so x+1 cant=0 and x-1 cant=0

so x cant= -1 nor 1

the domain would be x = all real numbers except -1 and 1

Answer 4

The domain of a function can be roughly defined as, "Whatever the x-value can be to make the function defined."

For example, take the equation "y = 2x"

x can be any value at all, and the y vlue will be defined.

However, if you take "y = 1/x", x can be any value except for 0 because 1/0 is undefined.

So in your case, f(x) = (x-1)/(x²-1)

Any fraction is undefined if the denominator is 0. So set your denominator (x²-1) equal to 0 and solve for x. This will give you any values of x that makes the denominator 0, thus making your fraciton undefined.

x²-1 = 0

(x+1)(x-1) = 0    [Factor - Difference of Squares]

x = {±1}

Therefore x cannot be equal to 1 or -1. In interval notation this is expressed as: (-infinitysign, -1) U (-1,1) U (1,infinitysign)

x can be anything from negative infinity to -1, from -1 to 1 [Meaning between -1 and 1], and 1 to infinity.

Answer 5

Factor the bottom

(x-1)/(x+1)(x-1)

=1/(x+1)