QuestionFind $f^{-1}(x)$ , and state the domain and range of $f^{-1}(x)$ for $f(x)=\sqrt{12}(x+5)$ .

Studdy Solution

STEP 1

Assumptions1. The function is $f(x)=\sqrt{12}(x+5)$ . We are looking for the inverse function, $f^{-1}(x)$ 3. We also need to find the domain and range of $f^{-1}(x)$

STEP 2

To find the inverse of a function, we switch $x$ and $y$ and then solve for $y$ . So, we start by replacing $f(x)$ with $y$ .
$y=\sqrt{12}(x+5)$

STEP 3

Next, we swap $x$ and $y$ .
$x=\sqrt{12}(y+5)$

STEP 4

Now we solve for $y$ to find $f^{-1}(x)$ . First, we divide both sides by $\sqrt{12}$ .
$\frac{x}{\sqrt{12}}=y+$

STEP 5

Then, we subtract5 from both sides to isolate $y$ .
$y=\frac{x}{\sqrt{12}}-5$

STEP 6

This is the expression for $f^{-1}(x)$ .
$f^{-1}(x)=\frac{x}{\sqrt{12}}-5$

STEP 7

Now we need to find the domain and range of $f^{-1}(x)$ . The domain of $f^{-1}(x)$ is the range of $f(x)$ , and the range of $f^{-1}(x)$ is the domain of $f(x)$ .

STEP 8

The domain of $f(x)$ is all real numbers, because $x$ can be any real number in the function $f(x)=\sqrt{12}(x+5)$ . Therefore, the range of $f^{-1}(x)$ is all real numbers.

STEP 9

The range of $f(x)$ is all real numbers greater than or equal to $-5\sqrt{12}$ , because the smallest value $f(x)$ can take is $-5\sqrt{12}$ when $x=$ . Therefore, the domain of $f^{-}(x)$ is all real numbers greater than or equal to $-5\sqrt{12}$ .
So, the inverse function is $f^{-}(x)=\frac{x}{\sqrt{12}}-5$ with domain $x \geq -5\sqrt{12}$ and range $y$ is all real numbers.

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