QuestionFind the inverse function $f^{-1}(x)$ and its domain and range for $f(x)=\sqrt{x}+4$ .

Studdy Solution

STEP 1

Assumptions1. The function is $f(x)=\sqrt{x}+4$ . We need to find the inverse function, $f^{-1}(x)$ 3. We also need to determine the domain and range of $f^{-1}(x)$

STEP 2

To find the inverse of a function, we first replace $f(x)$ with $y$ .
$y=\sqrt{x}+4$

STEP 3

Next, we switch the roles of $x$ and $y$ . This means we replace every $x$ in our equation with $y$ and every $y$ with $x$ .
$x=\sqrt{y}+$

STEP 4

Now, we solve this equation for $y$ to find $f^{-1}(x)$ . First, subtract4 from both sides.
$x-4=\sqrt{y}$

STEP 5

Then, square both sides to get rid of the square root.
$(x-4)^2=y$

STEP 6

Now we have found the inverse function, $f^{-1}(x)$ .
$f^{-1}(x)=(x-4)^2$

STEP 7

Next, we find the domain of $f^{-1}(x)$ . The domain of a function is the set of all possible input values (x-values) which will output real numbers.For the function $f^{-1}(x)=(x-4)^2$ , the domain is all real numbers because a square of any real number is a real number.
So, the domain of $f^{-1}(x)$ is $(-\infty, \infty)$ .

STEP 8

Finally, we find the range of $f^{-1}(x)$ . The range of a function is the set of all possible output values (y-values).For the function $f^{-1}(x)=(x-4)^2$ , the output can be any non-negative real number because a square of any real number is always non-negative.
So, the range of $f^{-1}(x)$ is $[0, \infty)$ .
The inverse function is $f^{-1}(x)=(x-4)^2$ , the domain of $f^{-1}(x)$ is $(-\infty, \infty)$ , and the range of $f^{-1}(x)$ is $[0, \infty)$ .

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