Question

Math

Posted 4 months ago

```
The one-to-one function $f$ is defined below.
$f(x)=\frac{2 x}{8-x}$
Find $f^{-1}(x)$, where $f^{-1}$ is the inverse of $f$.
Also state the domain and range of $f^{-1}$ in interval notation.
$f^{-1}(x)=$
Domain of $f^{-1}$ :
Range of $f^{-1}$ :
```

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Answer from Sia

Posted 4 months ago

Solution by Steps

step 1

To find the inverse function $f^{-1}(x)$, we start by setting $y = f(x)$. So, we have $y = \frac{2x}{8-x}$

step 2

Next, we solve for $x$ in terms of $y$. Multiply both sides by $(8-x)$ to get $y(8-x) = 2x$

step 3

Distribute $y$ on the left side: $8y - xy = 2x$

step 4

Rearrange the equation to isolate $x$: $8y = 2x + xy$

step 5

Factor out $x$ on the right side: $8y = x(2 + y)$

step 6

Solve for $x$: $x = \frac{8y}{2 + y}$

step 7

Replace $y$ with $x$ to get the inverse function: $f^{-1}(x) = \frac{8x}{2 + x}$

step 8

To find the domain of $f^{-1}(x)$, we need to determine the values of $x$ for which the function is defined. The denominator $2 + x$ must not be zero, so $x \neq -2$. Therefore, the domain of $f^{-1}(x)$ is $(-\infty, -2) \cup (-2, \infty)$

step 9

To find the range of $f^{-1}(x)$, we look at the domain of the original function $f(x)$. The function $f(x) = \frac{2x}{8-x}$ is defined for $x \neq 8$. As $x$ approaches $8$, $f(x)$ approaches infinity. Therefore, the range of $f(x)$ is $(-\infty, \infty)$. This means the range of $f^{-1}(x)$ is also $(-\infty, \infty)$

Answer

$f^{-1}(x) = \frac{8x}{2 + x}$

Domain of $f^{-1}(x)$: $(-\infty, -2) \cup (-2, \infty)$

Range of $f^{-1}(x)$: $(-\infty, \infty)$

Key Concept

Inverse Function

Explanation

To find the inverse of a function, solve for the input variable in terms of the output variable, then interchange the variables. The domain of the inverse function is the range of the original function, and vice versa.

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