Math / Algebra

QuestionFind a formula for $f^{-1}(x)$ . Identify the domain and range of $f^{-1}$ . Verify that $f$ and $f^{-1}$ are inverses. $f(x)=\sqrt{x+5}, x \geq-5$ $f^{-1}(x)=x^{2}-5$ (Simplify your answer.) Identify the domain of $f^{-1}$ . Select the correct choice below and, if necessary, fill in the answer box(es) to complete your A. $D=\{x \mid x \leq\}$ B. $D=\{x \mid \leq x \leq\}$ C. $D=\{x \mid x \neq\}$ D. $D=\{x \mid x \geq 0\}$ E. The domain is all real numbers.
Identify the range of $f^{-1}$ . Select the correct choice below and, if necessary, fill in the answer box(es) to complete your ch A. $R=\{y \mid \leq y \leq\}$ B. $R=\{y \mid y \leq \quad\}$ C. $R=\{y \mid y \geq-5\}$ D. $R=\{y \mid y \neq\}$ E. The range is all real numbers.
To verify that $f$ and $f^{-1}$ are inverses, determine $\left(f^{-1} \circ f\right)(x)$ and $\left(f \circ f^{-1}\right)(x)$ . $\left(f \circ f^{-1}\right)(\mathrm{x})=\square \quad\left(\mathrm{f}^{-1} \circ \mathrm{f}\right)(\mathrm{x})=\square$ (Simplify your answers.)

Studdy Solution

STEP 1

What is this asking? We're on a mission to find the inverse of a square root function, figure out where it lives (domain and range), and double-check that it's a true inverse! Watch out! Remember, the domain of the original function becomes the range of the inverse, and the range of the original function becomes the domain of the inverse.
Don't mix them up!

STEP 2

1. Find the Inverse
2. Determine the Domain and Range
3. Verify the Inverse

STEP 3

Let's begin with our function $f(x) = \sqrt{x+5}$ .
To find the inverse, we'll swap $x$ and $y$ , so we have $x = \sqrt{y+5}$ .

STEP 4

Now, let's solve for $y$ ! Square both sides to get rid of that pesky square root: $x^2 = y+5$ .

STEP 5

Finally, subtract 5 from both sides: $y = x^2 - 5$ .
So, our inverse function is $f^{-1}(x) = x^2 - 5$ !

STEP 6

The domain of $f(x)$ is given as $x \geq -5$ , and its range is $y \geq 0$ (since the square root of a number is always non-negative).

STEP 7

Remember, for inverse functions, the domain and range switch!
So, the domain of $f^{-1}(x)$ is $x \geq 0$ (which is D), and the range of $f^{-1}(x)$ is $y \geq -5$ (which is C).

STEP 8

Let's verify this by checking $(f^{-1} \circ f)(x)$ and $(f \circ f^{-1})(x)$ .

STEP 9

First, $(f^{-1} \circ f)(x) = f^{-1}(f(x)) = f^{-1}(\sqrt{x+5})$ . Substitute this into our inverse function: $(\sqrt{x+5})^2 - 5$ .
This simplifies to $x + 5 - 5 = x$ .
Awesome!

STEP 10

Next, $(f \circ f^{-1})(x) = f(f^{-1}(x)) = f(x^2 - 5)$ . Substitute this into our original function: $\sqrt{(x^2 - 5) + 5}$ .
This simplifies to $\sqrt{x^2}$ , which is $|x|$ .
Since the domain of $f^{-1}(x)$ is $x \geq 0$ , we know that $x$ is non-negative, so $|x| = x$ .
Perfect!

STEP 11

$f^{-1}(x) = x^2 - 5$ .
The domain of $f^{-1}$ is $D = \{x \mid x \geq 0\}$ .
The range of $f^{-1}$ is $R = \{y \mid y \geq -5\}$ . $(f \circ f^{-1})(x) = x$ and $(f^{-1} \circ f)(x) = x$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the Inverse
2. Determine the Domain and Range
3. Verify the Inverse

STEP 3

Let's begin with our function $f(x) = \sqrt{x+5}$ .
To find the inverse, we'll swap $x$ and $y$ , so we have $x = \sqrt{y+5}$ .

STEP 4

Now, let's solve for $y$ ! Square both sides to get rid of that pesky square root: $x^2 = y+5$ .

STEP 5

Finally, subtract 5 from both sides: $y = x^2 - 5$ .
So, our inverse function is $f^{-1}(x) = x^2 - 5$ !

STEP 6

The domain of $f(x)$ is given as $x \geq -5$ , and its range is $y \geq 0$ (since the square root of a number is always non-negative).

STEP 7

Remember, for inverse functions, the domain and range switch!
So, the domain of $f^{-1}(x)$ is $x \geq 0$ (which is D), and the range of $f^{-1}(x)$ is $y \geq -5$ (which is C).

STEP 8

Let's verify this by checking $(f^{-1} \circ f)(x)$ and $(f \circ f^{-1})(x)$ .

STEP 9

First, $(f^{-1} \circ f)(x) = f^{-1}(f(x)) = f^{-1}(\sqrt{x+5})$ . Substitute this into our inverse function: $(\sqrt{x+5})^2 - 5$ .
This simplifies to $x + 5 - 5 = x$ .
Awesome!

STEP 10

Next, $(f \circ f^{-1})(x) = f(f^{-1}(x)) = f(x^2 - 5)$ . Substitute this into our original function: $\sqrt{(x^2 - 5) + 5}$ .
This simplifies to $\sqrt{x^2}$ , which is $|x|$ .
Since the domain of $f^{-1}(x)$ is $x \geq 0$ , we know that $x$ is non-negative, so $|x| = x$ .
Perfect!

STEP 11

$f^{-1}(x) = x^2 - 5$ .
The domain of $f^{-1}$ is $D = \{x \mid x \geq 0\}$ .
The range of $f^{-1}$ is $R = \{y \mid y \geq -5\}$ . $(f \circ f^{-1})(x) = x$ and $(f^{-1} \circ f)(x) = x$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the Inverse2. Determine the Domain and Range3. Verify the Inverse

STEP 3

Let's **begin** with our function f(x)=x+5f(x) = \sqrt{x+5}f(x)=x+5​.To find the inverse, we'll **swap** xxx and yyy, so we have x=y+5x = \sqrt{y+5}x=y+5​.

STEP 4

Now, let's **solve** for yyy! **Square** both sides to get rid of that pesky square root: x2=y+5x^2 = y+5x2=y+5.

STEP 5

Finally, **subtract** 5 from both sides: y=x2−5y = x^2 - 5y=x2−5.So, our **inverse function** is f−1(x)=x2−5f^{-1}(x) = x^2 - 5f−1(x)=x2−5!

STEP 6

The **domain** of f(x)f(x)f(x) is given as x≥−5x \geq -5x≥−5, and its **range** is y≥0y \geq 0y≥0 (since the square root of a number is always non-negative).

STEP 7

Remember, for inverse functions, the domain and range switch!So, the **domain of** f−1(x)f^{-1}(x)f−1(x) is x≥0x \geq 0x≥0 (which is D), and the **range of** f−1(x)f^{-1}(x)f−1(x) is y≥−5y \geq -5y≥−5 (which is C).

STEP 8

Let's **verify** this by checking (f−1∘f)(x)(f^{-1} \circ f)(x)(f−1∘f)(x) and (f∘f−1)(x)(f \circ f^{-1})(x)(f∘f−1)(x).

STEP 9

First, (f−1∘f)(x)=f−1(f(x))=f−1(x+5)(f^{-1} \circ f)(x) = f^{-1}(f(x)) = f^{-1}(\sqrt{x+5})(f−1∘f)(x)=f−1(f(x))=f−1(x+5​). **Substitute** this into our inverse function: (x+5)2−5(\sqrt{x+5})^2 - 5(x+5​)2−5.This **simplifies** to x+5−5=xx + 5 - 5 = xx+5−5=x.Awesome!

STEP 10

STEP 11

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1. Find the Inverse
2. Determine the Domain and Range
3. Verify the Inverse

Let's begin with our function $f(x) = \sqrt{x+5}$ .
To find the inverse, we'll swap $x$ and $y$ , so we have $x = \sqrt{y+5}$ .

Now, let's solve for $y$ ! Square both sides to get rid of that pesky square root: $x^2 = y+5$ .

Finally, subtract 5 from both sides: $y = x^2 - 5$ .
So, our inverse function is $f^{-1}(x) = x^2 - 5$ !

The domain of $f(x)$ is given as $x \geq -5$ , and its range is $y \geq 0$ (since the square root of a number is always non-negative).

Remember, for inverse functions, the domain and range switch!
So, the domain of $f^{-1}(x)$ is $x \geq 0$ (which is D), and the range of $f^{-1}(x)$ is $y \geq -5$ (which is C).

Let's verify this by checking $(f^{-1} \circ f)(x)$ and $(f \circ f^{-1})(x)$ .

First, $(f^{-1} \circ f)(x) = f^{-1}(f(x)) = f^{-1}(\sqrt{x+5})$ . Substitute this into our inverse function: $(\sqrt{x+5})^2 - 5$ .
This simplifies to $x + 5 - 5 = x$ .
Awesome!