What is the inverse function of #f(x) = 2x + x^(1/2)#?

1 Answer
Jul 10, 2018

#color(blue)(y=(4x+1+sqrt(8x+1))/8#

#color(blue)(y=(4x+1-sqrt(8x+1))/8#

Explanation:

Exchange #x# and #y#:

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

Subtract #2y#:

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

Square both sides:

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

Expand #LHS#

#x^2-4xy+4y^2=y#

#x^2-4xy+4y^2-y=0#

#4y^2-(4x+1)y+x^2=0#

We now solve this for #y#, using the quadratic formula:

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

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

#color(blue)(y=(4x+1+sqrt(8x+1))/8#

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

#color(blue)(y=(4x+1-sqrt(8x+1))/8#