Question

Find the f^-1(x) and it’s domain

Answer 1

Answer:

$f^{-1}(x) = (x + 8)^2$

$x \ge -8$

Step-by-step explanation:

Given

$f(x) = \sqrt x - 8$

Solving (a): $f^{-1}(x)$

We have:

$f(x) = \sqrt x - 8$

Express f(x) as y

$y = \sqrt x - 8$

Swap x and y

$x = \sqrt y - 8$

Add 8 to $both\ sides$

$x + 8 = \sqrt y - 8 + 8$

$x + 8 = \sqrt y$

Square both sides

$(x + 8)^2 = y$

Rewrite as:

$y = (x + 8)^2$

Express y as: $f^{-1}(x)$

$f^{-1}(x) = (x + 8)^2$

To determine the domain, we have:

The original function is $f(x) = \sqrt x - 8$

The range of this is: $f(x) \ge -8$

The $domain$ of the $inverse$ function is the $range$ of the $original$ function.

Hence, the domain is:

$x \ge -8$