Question

I need help with number 4

Answer 1

Answer:  Choice D

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

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Explanation:

First we replace f(x) with y. This is because both y and f(x) are outputs of a function.

To find the inverse, we swap x and y and solve for y like so

$y = \sqrt{x-2}+3\\\\x = \sqrt{y-2}+3 \ \text{ .... swap x and y; isolate y}\\\\x-3 = \sqrt{y-2}\\\\(x-3)^2 = y-2 \ \text{ ... square both sides}\\\\(x-3)^2+2 = y\\\\y = (x-3)^2+2\\\\f^{-1}(x) = (x-3)^2+2$

Note: because the range of the original function is $y \ge 3$, this means the domain of the inverse is $x \ge 3$. The domain and range swap roles because of the swap of x and y.

As the graph shows below, the original and its inverse are symmetrical about the mirror line y = x. One curve is the mirror image of the other over this dashed line.

Answer 2

Answer:

D

Step-by-step explanation:

let y = f(x) and rearrange making x the subject

y = $\sqrt{x-2}$ + 3 ( subtract 3 from both sides )

y - 3 = $\sqrt{x-2}$ ( square both sides )

(y - 3)² = x - 2 ( add 2 to both sides )

(y - 3)² + 2 = x

Change y back into terms of x with x = $f^{-1}$ (x) , then

$f^{-1}$ (x) = (x - 3)² + 2 → D