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

1 Answer
May 9, 2016

Answer:

Use an algebraic method to find range is #(0, oo)#

Explanation:

If a particular value of #y# is in the range, then we can solve for #x# to give that value of #y#.

Suppose:

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

Adding #x# to both sides we get:

#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 #x^2+y^2# from both sides to get:

#2xy = 1-y^2#

Note that we must have #y != 0# since otherwise this equation would become #0 = 1#. So in particular #y=0# is not in the range.

Divide both sides by #2y# to get:

#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 #y > 0# then #abs(y) = y# and this becomes:

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

If #y < 0# then #abs(y) = -y# and this becomes:

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

So there is a solution for #x# if and only if #y > 0#.

So the range is #(0, oo)#