Math / Algebra

QuestionA formula for a function $y=f(x)$ is $f(x)=10 x^{5}$ . Find $f^{-1}(x)$ and identify the domain and range of $f^{-1}(x)$ . To check the answer, determine whether $f\left(f^{-1}(x)\right)=f^{-1}(f(x))=x$ . $f^{-1}(x)=$ $\square$ (Type an exact answer, using radicals as needed.)

Studdy Solution

STEP 1

1. We are given the function $f(x) = 10x^5$ .
2. We need to find the inverse function $f^{-1}(x)$ .
3. We need to identify the domain and range of $f^{-1}(x)$ .
4. We need to verify that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$ .

STEP 2

1. Express $y = f(x)$ in terms of $x$ and solve for $x$ in terms of $y$ .
2. Determine the domain and range of $f(x)$ .
3. Identify the domain and range of $f^{-1}(x)$ .
4. Verify that $f(f^{-1}(x)) = x$ .
5. Verify that $f^{-1}(f(x)) = x$ .

STEP 3

Express $y = f(x)$ in terms of $x$ :
$y = 10x^5$
Solve for $x$ in terms of $y$ :
$x^5 = \frac{y}{10}$
Take the fifth root of both sides:
$x = \left(\frac{y}{10}\right)^{1/5}$
Thus, the inverse function is:
$f^{-1}(x) = \left(\frac{x}{10}\right)^{1/5}$

STEP 4

Determine the domain and range of $f(x) = 10x^5$ :
- The domain of $f(x)$ is all real numbers, $\mathbb{R}$ . - Since $x^5$ is an odd power function, the range of $f(x)$ is also all real numbers, $\mathbb{R}$ .

STEP 5

Identify the domain and range of $f^{-1}(x)$ :
- The domain of $f^{-1}(x)$ is the range of $f(x)$ , which is $\mathbb{R}$ . - The range of $f^{-1}(x)$ is the domain of $f(x)$ , which is $\mathbb{R}$ .

STEP 6

Verify that $f(f^{-1}(x)) = x$ :
$f(f^{-1}(x)) = f\left(\left(\frac{x}{10}\right)^{1/5}\right)$
$= 10\left(\left(\frac{x}{10}\right)^{1/5}\right)^5$
$= 10 \cdot \frac{x}{10}$
$= x$

STEP 7

Verify that $f^{-1}(f(x)) = x$ :
$f^{-1}(f(x)) = f^{-1}(10x^5)$
$= \left(\frac{10x^5}{10}\right)^{1/5}$
$= (x^5)^{1/5}$
$= x$
The inverse function is:
$f^{-1}(x) = \left(\frac{x}{10}\right)^{1/5}$
The domain and range of $f^{-1}(x)$ are both $\mathbb{R}$ .

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QuestionA formula for a function $y=f(x)$ is $f(x)=10 x^{5}$ . Find $f^{-1}(x)$ and identify the domain and range of $f^{-1}(x)$ . To check the answer, determine whether $f\left(f^{-1}(x)\right)=f^{-1}(f(x))=x$ . $f^{-1}(x)=$ $\square$ (Type an exact answer, using radicals as needed.)

Studdy Solution

STEP 1

1. We are given the function $f(x) = 10x^5$ .
2. We need to find the inverse function $f^{-1}(x)$ .
3. We need to identify the domain and range of $f^{-1}(x)$ .
4. We need to verify that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$ .

STEP 2

1. Express $y = f(x)$ in terms of $x$ and solve for $x$ in terms of $y$ .
2. Determine the domain and range of $f(x)$ .
3. Identify the domain and range of $f^{-1}(x)$ .
4. Verify that $f(f^{-1}(x)) = x$ .
5. Verify that $f^{-1}(f(x)) = x$ .

STEP 3

Express $y = f(x)$ in terms of $x$ :
$y = 10x^5$
Solve for $x$ in terms of $y$ :
$x^5 = \frac{y}{10}$
Take the fifth root of both sides:
$x = \left(\frac{y}{10}\right)^{1/5}$
Thus, the inverse function is:
$f^{-1}(x) = \left(\frac{x}{10}\right)^{1/5}$

STEP 4

Determine the domain and range of $f(x) = 10x^5$ :
- The domain of $f(x)$ is all real numbers, $\mathbb{R}$ . - Since $x^5$ is an odd power function, the range of $f(x)$ is also all real numbers, $\mathbb{R}$ .

STEP 5

Identify the domain and range of $f^{-1}(x)$ :
- The domain of $f^{-1}(x)$ is the range of $f(x)$ , which is $\mathbb{R}$ . - The range of $f^{-1}(x)$ is the domain of $f(x)$ , which is $\mathbb{R}$ .

STEP 6

Verify that $f(f^{-1}(x)) = x$ :
$f(f^{-1}(x)) = f\left(\left(\frac{x}{10}\right)^{1/5}\right)$
$= 10\left(\left(\frac{x}{10}\right)^{1/5}\right)^5$
$= 10 \cdot \frac{x}{10}$
$= x$

STEP 7

Verify that $f^{-1}(f(x)) = x$ :
$f^{-1}(f(x)) = f^{-1}(10x^5)$
$= \left(\frac{10x^5}{10}\right)^{1/5}$
$= (x^5)^{1/5}$
$= x$
The inverse function is:
$f^{-1}(x) = \left(\frac{x}{10}\right)^{1/5}$
The domain and range of $f^{-1}(x)$ are both $\mathbb{R}$ .

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Studdy Solution

STEP 1

STEP 2

STEP 3

STEP 4

Determine the domain and range of f(x)=10x5 f(x) = 10x^5 f(x)=10x5:- The domain of f(x) f(x) f(x) is all real numbers, R \mathbb{R} R. - Since x5 x^5 x5 is an odd power function, the range of f(x) f(x) f(x) is also all real numbers, R \mathbb{R} R.

STEP 5

Identify the domain and range of f−1(x) f^{-1}(x) f−1(x):- The domain of f−1(x) f^{-1}(x) f−1(x) is the range of f(x) f(x) f(x), which is R \mathbb{R} R. - The range of f−1(x) f^{-1}(x) f−1(x) is the domain of f(x) f(x) f(x), which is R \mathbb{R} R.

STEP 6

STEP 7

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Determine the domain and range of $f(x) = 10x^5$ :
- The domain of $f(x)$ is all real numbers, $\mathbb{R}$ . - Since $x^5$ is an odd power function, the range of $f(x)$ is also all real numbers, $\mathbb{R}$ .

Identify the domain and range of $f^{-1}(x)$ :
- The domain of $f^{-1}(x)$ is the range of $f(x)$ , which is $\mathbb{R}$ . - The range of $f^{-1}(x)$ is the domain of $f(x)$ , which is $\mathbb{R}$ .