Respuesta :

gmany

Answer:

[tex]\huge\boxed{g^{-1}(x)=\sqrt{x-1}\ \text{for}\ x\geq1}[/tex]

Step-by-step explanation:

[tex]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=\sqrt{x-1}\\\\\text{Domain}\\\\x-1\geq0\qquad|\text{add 1 to both sides}\\\\x-1+1\geq0+1\\\\x\geq1[/tex]

loneguy

[tex]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 \sqrt{x-1}\\\\\\\implies g^{-1} (x) = \pm\sqrt{x-1}[/tex]

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