Question

Identify the inverse function of f(x) = VX - 2 + 3.

Answer 1

Answer:

$\huge\boxed{f^{-1}(x) = (x-3)^2+2}$

Step-by-step explanation:

$f(x) = \sqrt{x-2} + 3$

Replace y = f(x)

$y = \sqrt{x-2} + 3$

Exchange x and y

$x = \sqrt{y-2}+3$

Solve for y

$x = \sqrt{y-2}+3$

Subtracting both sides by 3

$x - 3 = \sqrt{y-2}$

Taking square on both sides

$(x-3)^2 = y -2$

Adding 2 to both sides

$y = (x-3)^2+2$

Substitute y = $f^{-1}(x)$

$f^{-1}(x) = (x-3)^2+2$

Answer 2

Answer:

$\boxed{ {f}^{ - 1} (x) = {(x - 3)}^{2} + 2}$

Option D is the correct option

Step-by-step explanation:

$\mathsf{f(x) = \sqrt{x - 2} + 3}$

Replace f(x) with y

$\mathsf{y = \sqrt{x - 2} + 3}$

Interchange variables

$\mathsf{x = \sqrt{y - 2} + 3}$

$\mathsf{{(x - 3)}^{2} = {( \sqrt{y - 2)} }^{2} }$

$\mathsf{ {(x - 3)}^{2} = y - 2}$

$\mathsf{ y = {(x - 3)}^{2} + 2}$

Replace y with f ⁻¹( x )

$\mathsf{ {f}^{ - 1} (x) = {(x - 3)}^{2} + 2}$

Hope I helped!

Best regards!