The domain of the function, f(x) = 1/ sqroot (X^2-X-2 )?

Answer 1

The domain of a function like this, means all the values of X for which the function makes sense. In this case, that means all the values of X, except for the value that makes X^2 - X -2 = 0. That's because you can't divide by zero.

So what value of X makes X^2 - X -2 = 0 ?
Use the solution for a quadratic equation, with a = 1, b = -1, and c = -2
roots = [-b +/- sqrt( b^2 - 4ac) ] / (2a)
roots = [1 +/- sqrt( 1 + 8) ] / (2)
roots = [1 +/- sqrt( 9 ) ] / 2
roots = [1 +/- 3 ] / 2
roots = 4/2 = 2, and -2/2 = -1

So the domain of the function f(x) is all numbers, except x = 2 and x = -1.

Answer 2

The denominator can be written as (factoring)

sqrt[(x-2)(x+1)] and the function will be undefined for real values if this denominator is zero or negative.

It is negative for x = -1 and 2 so those are not in the domain.

Solve the inequality (x-2)(x+1)>0 or test the sign of each factor on the intervals - infinity to -1, -1 to 2, and 2 to positive infinity to find that the domain is (-infinity,-1) union (2, +infinity)

Answer 3

the domain = (-inf,-1)U(2, inf) because if u take,for example, x=0 than sqroot(x^2-x-2) will be less than 0 (0^2-0-2=-2)and u cannot have a negative number under sqroot,except of complex numbers but we r talking about real numbers.
so the domain is (-infinite,-1)U(2,infinite)

Answer 4

f(x) = 1 / √(x^2 -x -2)

1) because it is a fraction, you have to check if there are values of x that would give a denominator of zero (1/0 is not defined).

2) because the function is a square root, you have to check if there are values of x that result in the quadratic (x^2 -x -2) being less than 0 -- you cannot have the square root of a negative number.

The domain could be anything that is not contained in 1) and 2).

Answer 5

the respond would be (a) indoors the least complicated variety, the question is finding you find the values of Y on the comparable time as x is 0 or a million or 2 or 3. so as which you hav to swap each and every and each x-fee into the Y=3x-5 expression to be certain for one among those y... i.e on the comparable time as x=0 y=3(0)-5=-5 on the comparable time as x=a million y=3(a million)-5=-2 on the comparable time as x=2 y=3(2)-5=a million on the comparable time as x=3 y=3(3)-5=4

Answer 6

x^2-x-2=(x-2)(x+1)
f(x)=1/sqrt[(x-2)(x+1)]
Because we have a square root,
(x-2)(x+1) >=0
x> 2 and x> -1: Both imply x >=2
Domain consists of all numbers >=2.

Answer 7

x^2 -x -2 > 0

roots 2, -1

(x-2)(x+1) > 0

then x>2 or x<-1

Domain = (-inf,-1) U (2, inf)