To find the inverse [tex]\( f^{-1}(x) \)[/tex] of the function [tex]\( f(x) = \sqrt{4 - x} + 6 \)[/tex] and its domain, let's follow these steps:
1. Express the function in terms of [tex]\( y \)[/tex]:
[tex]\[
y = \sqrt{4 - x} + 6
\][/tex]
2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[
y - 6 = \sqrt{4 - x}
\][/tex]
3. Square both sides to eliminate the square root:
[tex]\[
(y - 6)^2 = 4 - x
\][/tex]
4. Isolate [tex]\( x \)[/tex]:
[tex]\[
x = 4 - (y - 6)^2
\][/tex]
5. Rewrite the expression with [tex]\( x \)[/tex] as the dependent variable:
[tex]\[
f^{-1}(x) = 4 - (x - 6)^2
\][/tex]
Now, we have found the inverse function:
[tex]\[
f^{-1}(x) = 4 - (x - 6)^2
\][/tex]
Next, we need to determine the domain of [tex]\( f^{-1}(x) \)[/tex]. The domain of [tex]\( f(x) \)[/tex] was given as [tex]\( (-\infty, 4] \)[/tex]. The range of [tex]\( f(x) \)[/tex] will be the domain of [tex]\( f^{-1}(x) \)[/tex]:
- When [tex]\( x = 4 \)[/tex], [tex]\( f(4) = \sqrt{4-4} + 6 = 6 \)[/tex].
- As [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex], [tex]\( \sqrt{4-x} \)[/tex] becomes very large. Hence [tex]\( f(x) \)[/tex] becomes very large too, approaching [tex]\( +\infty \)[/tex].
Therefore, the range of [tex]\( f(x) \)[/tex] is [tex]\([6, +\infty)\)[/tex], which means the domain of [tex]\( f^{-1}(x) \)[/tex] is also [tex]\([6, +\infty)\)[/tex].
Hence, the inverse function is:
[tex]\[
f^{-1}(x) = 4 - (x - 6)^2
\][/tex]
And the domain of [tex]\( f^{-1} \)[/tex] in interval notation is:
[tex]\[
[6, +\infty)
\][/tex]
