Find the domain and range of f(x) = sqrt x / x^2 + x - 2?

Answer 1

well KG
▪Domain . we have to simplify this fraction;

▪ Step 1

f(x) = √((x) /( x^2 + x - 2))

Domain (x) = (-∞, +∞)
and range (x) = R ,Or (-∞, +∞)
{ explanation; R – the set of real numbers,}

▪ Step 2
Domain of ( x^2 + x - 2) , first we need to find the roots of this function.

we have to find Two numbers like 'a' & 'b' that when we add them ,the result would be ' -2 ' and we multiply them ,the result would be ' +1' ; so we have ;
( x^2 + x - 2) = ( x-1)( x+2)
if f ( x^2 + x - 2) = 0
so
( x-1)( x+2) = 0 ;
( x-1)=0 ; x=+1
Or
( x+2) = 0 ; x= -2;
● Notice : as you know if we put '+1' in this function ( x-1)( x+2) so we have ( +1-1)(x+2) = 0 * (x+2) = 0 ,you see he reault is 0.or if you put '-2' in this function ( x-1)( x+2) so we have ( x-1)(-2+2) =(x-1) * 0 = 0, you see he reault is 0 AGAIN. THE PROBLEM IS ; this function is in the Denominator of our fraction ,and as you know ' Zero ' cant be there. so
domain of x^2 + x - 2 = all porsitive and negative numbers EXCEP zero , ' +1' & '-2'
Or
Domain of x^2 + x - 2 =R - { -2 , Zero ,+1}

▪ Step 3

Now domain of whole of fraction of f(x)
well as you Negative numbers cant be in " √ " but Zero and postive numbers could be.
and (x / x^2 + x - 2) > 0 , and why not ≥0 ?? coz earlier we did prove in the Denominator of any fraction ' Zero ' cant be there.
x > (x^2 + x - 2) * 0
x > 0
Domain of (x) = [ -∞ , + ∞)
Domain of (x^2 + x - 2) =R - { -2 , Zero ,+1}

Here we should get an intersection of Domain of ( x) and Domain of (x^2 + x - 2); but with this conditioan that " x > 0"
Therefor ;
Domain (fx) = [+ 2 ,+ ∞)

▪ Step 4
now Rnage its very interesting

just put afew number in the f(x) i would do that now
if x =+ 2 ; f(x) = √2/ ( 2-1)(2+2) = √2/4 = √1/2 = 0.70
if x =+ 3 ; f(x) = √3/ ( 3-1)(3+2) = √3/10 = 0.54
if x =+ 4 ; f(x) = √4/ ( 4-1)(4+2) = √4/18 = 0.47
you see any time ' x 'gets bigger ' y ' would get smaller
so ;
Range f(x) = (0 , +1) ,the range is just between 0 and 1 but it never be 0 or 1 .

Its you correct answer,
Good luck KG .

Answer 2

the area of f(x) is all reals or ( - ?, ?) the form of f(x) [4, ?) or y ? 4 WHY? any actual quantity could be used as enter into f(x). there is not any quantity which will f(x) undefined. For the style, x^2 will continuously be ? 0, for any fee of x. so the minimum fee would be at x = 0, f(x) = 4 y ? 4 or the style = [ 4, ?) For g(x) the area must be values of x the place (3x - 5) ? 0 as we don't take ? of unfavourable numbers. 3x - 5 ? 0 3x ? 5 x ? 5/3 or [5/3, ?) for the form of g(x), it will be 0 and all valuable numbers y ? 0 or [ 0, ?) For (fog)(x) you began out super, yet why did you placed the expression = 0? (fog) (x) = 3x - a million and in basic terms bypass away it at that. it fairly is the equation of a line this permits the area to be all actual numbers --> (-?, ?) And the style will additionally be all actual numbers --> (-?, ?)

Answer 3

to get the domain of sqrt( x/x^2 + x-2)
we make : x/x^2 + x-2 = 0
multiply the equation by x
1 + x^2 - 2x = 0
x^2 -2x + 1 = 0
(x-1)^2 = 0 ... the zero of the equation is x=1
by checking the sign of the function .. we find that (1/x + x - 2) is positive where x>=1 ... that's the domain
the range is [0,+infinity)

Answer 4

If U mean sqrt(x/(x^2+x-2))
to get its domain :
1) Put x>= 0 & (x^2 + x -2 > 0),
get the intersection of the two solution intervals which give
]1 , infinty [
2) Put x<= 0 & (x^2 + x -2 < 0),
get the intersection of the two solution intervals which give
]-2 , 0 [
So the domain is ] -2 , 0 [ U ]1 , infinty [
The range is ] 0 , infinty [

Answer 5

Easy. It is "2"

Answer 6

1) [0,1[ U ]1;...[ if f(x) = sqrt(x) / (x^2+x-2)
2) ]0, ...[ if f(x) = x - 2 + sqrt(x) / x^2