409,179 views
2 votes
2 votes
G(x)= x^2+1 I need help finding the inverse function to this. work must be shown

User Mike Davis
by
2.5k points

2 Answers

19 votes
19 votes

Answer:


\huge\boxed{g^(-1)(x)=√(x-1)\ \text{for}\ x\geq1}

Explanation:


g(x)=x^2+1\to y=x^2+1\\\\\text{exchange x with y and vice versa}\\\\x=y^2+1\\\\\text{solve for}\ y\\\\y^2+1=x\qquad|\text{subtract 1 from both sides}\\\\y^2=x-1\to y=√(x-1)\\\\\text{Domain}\\\\x-1\geq0\qquad|\text{add 1 to both sides}\\\\x-1+1\geq0+1\\\\x\geq1

User NilsH
by
2.6k points
14 votes
14 votes


g(x) = x^2+1\\\\\text{Write g(x) as}~ y = x^2 +1\\\\\text{Replace x with y: }\\\\x = y^2 +1\\\\\text{Solve for y:}\\\\x = y^2 + 1 \\\\\implies y = \pm √(x-1)\\\\\\\implies g^(-1) (x) = \pm√(x-1)

User Chocolata
by
3.0k points