# What is the range of the function #y = sqrt(x^2+1)-x# ?

##### 1 Answer

Use an algebraic method to find range is

#### Explanation:

If a particular value of

Suppose:

#y = sqrt(x^2+1)-x#

Adding

#y+x = sqrt(x^2+1)#

Squaring both sides (which may introduce spurious solutions) we get:

#y^2+2xy+x^2 = x^2+1#

Subtract

#2xy = 1-y^2#

Note that we must have

Divide both sides by

#x = (1-y^2)/(2y)#

So this is a requirement. To see if it is sufficient, substitute it back in the left hand side of the original equation:

#sqrt(x^2+1)-x#

#= sqrt(((1-y^2)/(2y))^2+1)-(1-y^2)/(2y)#

#= sqrt(((1-2y^2+y^4)+4y^2)/(4y^2))-(1-y^2)/(2y)#

#=sqrt((1+y^2)^2/(4y^2))-(1-y^2)/(2y)#

#=(1+y^2)/(2abs(y))-(1-y^2)/(2y)#

If

#(1+y^2)/(2y)-(1-y^2)/(2y) = (2y^2)/(2y) = y#

If

#-(1+y^2)/(2y)-(1-y^2)/(2y) = -2/(2y) = -1/y != y#

So there is a solution for

So the range is