Math / Algebra

Question(2 points) Let $f(x)=7+\sqrt{x-9}$ . Then find each of the following, giving all domain (a) $f^{-1}(x)=$ $\square$ (b) The domain of $f$ is $\square$ (c) The domain of $f^{-1}$ is $\square$ (d) The range of $f$ is $\square$ (e) The range of $f^{-1}$ is $\square$

Studdy Solution

STEP 1

1. We are given the function $f(x) = 7 + \sqrt{x-9}$ .
2. We need to find the inverse function $f^{-1}(x)$ .
3. We need to determine the domain and range of both $f$ and $f^{-1}$ .

STEP 2

1. Determine the domain of $f(x)$ .
2. Find the inverse function $f^{-1}(x)$ .
3. Determine the domain of $f^{-1}(x)$ .
4. Determine the range of $f(x)$ .
5. Determine the range of $f^{-1}(x)$ .

STEP 3

To find the domain of $f(x) = 7 + \sqrt{x-9}$ , we need the expression under the square root to be non-negative:
$x - 9 \geq 0$
Solve for $x$ :
$x \geq 9$
Thus, the domain of $f$ is:
$[9, \infty)$

STEP 4

To find the inverse function $f^{-1}(x)$ , start by setting $y = f(x)$ :
$y = 7 + \sqrt{x-9}$
Solve for $x$ in terms of $y$ :
$y - 7 = \sqrt{x-9}$
Square both sides:
$(y - 7)^2 = x - 9$
Solve for $x$ :
$x = (y - 7)^2 + 9$
Thus, the inverse function is:
$f^{-1}(x) = (x - 7)^2 + 9$

STEP 5

The domain of $f^{-1}(x)$ is the range of $f(x)$ . From the expression for $f(x) = 7 + \sqrt{x-9}$ , the smallest value occurs when $x = 9$ , giving:
$f(9) = 7 + \sqrt{9-9} = 7$
Since the square root function can take any non-negative value, the range of $f(x)$ is:
$[7, \infty)$
Thus, the domain of $f^{-1}$ is:
$[7, \infty)$

STEP 6

The range of $f(x)$ was determined in the previous step as:
$[7, \infty)$

STEP 7

The range of $f^{-1}(x)$ is the domain of $f(x)$ , which is:
$[9, \infty)$
The solutions are: (a) $f^{-1}(x) = (x - 7)^2 + 9$ (b) The domain of $f$ is $[9, \infty)$ (c) The domain of $f^{-1}$ is $[7, \infty)$ (d) The range of $f$ is $[7, \infty)$ (e) The range of $f^{-1}$ is $[9, \infty)$

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Question(2 points) Let $f(x)=7+\sqrt{x-9}$ . Then find each of the following, giving all domain (a) $f^{-1}(x)=$ $\square$ (b) The domain of $f$ is $\square$ (c) The domain of $f^{-1}$ is $\square$ (d) The range of $f$ is $\square$ (e) The range of $f^{-1}$ is $\square$

Studdy Solution

STEP 1

1. We are given the function $f(x) = 7 + \sqrt{x-9}$ .
2. We need to find the inverse function $f^{-1}(x)$ .
3. We need to determine the domain and range of both $f$ and $f^{-1}$ .

STEP 2

1. Determine the domain of $f(x)$ .
2. Find the inverse function $f^{-1}(x)$ .
3. Determine the domain of $f^{-1}(x)$ .
4. Determine the range of $f(x)$ .
5. Determine the range of $f^{-1}(x)$ .

STEP 3

To find the domain of $f(x) = 7 + \sqrt{x-9}$ , we need the expression under the square root to be non-negative:
$x - 9 \geq 0$
Solve for $x$ :
$x \geq 9$
Thus, the domain of $f$ is:
$[9, \infty)$

STEP 4

To find the inverse function $f^{-1}(x)$ , start by setting $y = f(x)$ :
$y = 7 + \sqrt{x-9}$
Solve for $x$ in terms of $y$ :
$y - 7 = \sqrt{x-9}$
Square both sides:
$(y - 7)^2 = x - 9$
Solve for $x$ :
$x = (y - 7)^2 + 9$
Thus, the inverse function is:
$f^{-1}(x) = (x - 7)^2 + 9$

STEP 5

STEP 6

The range of $f(x)$ was determined in the previous step as:
$[7, \infty)$

STEP 7

The range of $f^{-1}(x)$ is the domain of $f(x)$ , which is:
$[9, \infty)$
The solutions are: (a) $f^{-1}(x) = (x - 7)^2 + 9$ (b) The domain of $f$ is $[9, \infty)$ (c) The domain of $f^{-1}$ is $[7, \infty)$ (d) The range of $f$ is $[7, \infty)$ (e) The range of $f^{-1}$ is $[9, \infty)$

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Studdy Solution

STEP 1

1. We are given the function f(x)=7+x−9 f(x) = 7 + \sqrt{x-9} f(x)=7+x−9​.2. We need to find the inverse function f−1(x) f^{-1}(x) f−1(x).3. We need to determine the domain and range of both f f f and f−1 f^{-1} f−1.

STEP 2

1. Determine the domain of f(x) f(x) f(x).2. Find the inverse function f−1(x) f^{-1}(x) f−1(x).3. Determine the domain of f−1(x) f^{-1}(x) f−1(x).4. Determine the range of f(x) f(x) f(x).5. Determine the range of f−1(x) f^{-1}(x) f−1(x).

STEP 3

To find the domain of f(x)=7+x−9 f(x) = 7 + \sqrt{x-9} f(x)=7+x−9​, we need the expression under the square root to be non-negative:x−9≥0 x - 9 \geq 0 x−9≥0Solve for x x x:x≥9 x \geq 9 x≥9Thus, the domain of f f f is:[9,∞) [9, \infty) [9,∞)

STEP 4

STEP 5

STEP 6

The range of f(x) f(x) f(x) was determined in the previous step as:[7,∞) [7, \infty) [7,∞)

STEP 7

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1. We are given the function $f(x) = 7 + \sqrt{x-9}$ .
2. We need to find the inverse function $f^{-1}(x)$ .
3. We need to determine the domain and range of both $f$ and $f^{-1}$ .

1. Determine the domain of $f(x)$ .
2. Find the inverse function $f^{-1}(x)$ .
3. Determine the domain of $f^{-1}(x)$ .
4. Determine the range of $f(x)$ .
5. Determine the range of $f^{-1}(x)$ .

To find the domain of $f(x) = 7 + \sqrt{x-9}$ , we need the expression under the square root to be non-negative:
$x - 9 \geq 0$
Solve for $x$ :
$x \geq 9$
Thus, the domain of $f$ is:
$[9, \infty)$

The range of $f(x)$ was determined in the previous step as:
$[7, \infty)$