The one-to-one function $f$ is defined below.
$$
f(x)=\frac{x-1}{5 x+6}
$$

Find $f^{-1}(x)$, where $f^{-1}$ is the inverse of $f$.
Also state the domain and range of $f^{-1}$ in interval notation.
$$
f^{-1}(x)=\frac{6 x+1}{1-5 x}
$$

Domain of $f^{-1}: \square$

Range of $f^{-1}: \square$

Question

The one-to-one function $f$ is defined below.
$$
f(x)=\frac{x-1}{5 x+6}
$$

Find $f^{-1}(x)$, where $f^{-1}$ is the inverse of $f$.
Also state the domain and range of $f^{-1}$ in interval notation.
$$
f^{-1}(x)=\frac{6 x+1}{1-5 x}
$$

Domain of $f^{-1}: \square$

Range of $f^{-1}: \square$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the inverse function $f^{-1}(x)$, we start with the given function $f(x) = \frac{x-1}{5x+6}$. ‖ ‖ step 2 ⋮ Set $y = f(x)$, so $y = \frac{x-1}{5x+6}$. ‖ ‖ step 3 ⋮ Solve for $x$ in terms of $y$: $y(5x + 6) = x - 1$. ‖ ‖ step 4 ⋮ Distribute $y$: $5xy + 6y = x - 1$. ‖ ‖ step 5 ⋮ Rearrange to isolate $x$: $5xy - x = -1 - 6y$. ‖ ‖ step 6 ⋮ Factor out $x$: $x(5y - 1) = -1 - 6y$. ‖ ‖ step 7 ⋮ Solve for $x$: $x = \frac{-1 - 6y}{5y - 1}$. ‖ ‖ step 8 ⋮ Therefore, the inverse function is $f^{-1}(x) = \frac{6x + 1}{1 - 5x}$. ‖ ‖ step 9 ⋮ To find the domain of $f^{-1}(x)$, we need to ensure the denominator is not zero: $1 - 5x eq 0$. ‖ ‖ step 10 ⋮ Solve for $x$: $x eq \frac{1}{5}$. ‖ ‖ step 11 ⋮ Thus, the domain of $f^{-1}(x)$ is $(-\infty, \frac{1}{5}) \cup (\frac{1}{5}, \infty)$. ‖ ‖ step 12 ⋮ To find the range of $f^{-1}(x)$, we need to consider the domain of the original function $f(x)$. ‖ ‖ step 13 ⋮ The domain of $f(x)$ is all real numbers except where the denominator is zero: $5x + 6 eq 0$. ‖ ‖ step 14 ⋮ Solve for $x$: $x eq -\frac{6}{5}$. ‖ ‖ step 15 ⋮ Therefore, the range of $f^{-1}(x)$ is $(-\infty, -\frac{6}{5}) \cup (-\frac{6}{5}, \infty)$. ‖ ∻Answer∻ ⚹ Domain of $f^{-1}(x)$: $(-\infty, \frac{1}{5}) \cup (\frac{1}{5}, \infty)$ ⚹ ⚹ Range of $f^{-1}(x)$: $(-\infty, -\frac{6}{5}) \cup (-\frac{6}{5}, \infty)$ ⚹ ∻Key Concept∻ ⚹Inverse Function⚹ ∻Explanation∻ ⚹To find the inverse of a function, solve for $x$ in terms of $y$ and then interchange $x$ and $y$. The domain of the inverse function is the range of the original function, and vice versa.⚹◊What is the process for finding the domain and range of the inverse of a one-to-one function?⍭ Generate me a similar question◊

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