91Ó°ÊÓ

An inverse function, denoted as \( f^{-1}(x) \), is a function that reverses the operation of the original function \( f(x) \). When you apply \( f^{-1} \) to the output of \( f \), it returns you back to the input. To find the inverse of a given function, such as \( f(x) = \frac{x}{1-3x} \), you start by swapping the roles of \( x \) and \( y \).
Here's a simple process to find an inverse:

• Begin with the equation taking \( y = f(x) \).
• Swap \( x \) and \( y \) to get \( x = f(y) \).
• Solve this new equation for \( y \), which gives you \( f^{-1}(x) \).

For our example, after swapping and solving, we find that \( f^{-1}(x) = \frac{x}{1+3x} \). Finding an inverse is like rewinding a process to see what points or values began the function, and this new equation represents exactly that.

The horizontal line test is an easy way to determine if a function is one-to-one. A function is one-to-one if each input produces a unique output. Graphically, this means a horizontal line should intersect the function's graph at most one time anywhere along the function.If you draw a horizontal line across the graph of \( f(x) = \frac{x}{1-3x} \) and it touches the graph at more than one point along its path, then the function is not one-to-one. However, for this function, any horizontal line will only meet the graph once.

• A one-to-one function will pass the horizontal line test.
• This test helps confirm that a function has an inverse.

Since no horizontal line hits more than one spot, and using the algebraic check \( f(a) = f(b) \) simplifies to \( a = b \), we confirm \( f(x) \) is indeed one-to-one. This guarantees the existence of an inverse for \( f(x) \).

The domain of a function consists of all the possible input values \( x \), while the range consists of all possible output values \( y \). When finding an inverse function, it's important to switch the domain and range between the function and its inverse.For \( f(x) = \frac{x}{1-3x} \), to ensure the function is well-defined, \( x \) cannot be \( \frac{1}{3} \) because it would make the denominator zero. Therefore, the domain is all real numbers except \( \frac{1}{3} \).

• The domain of \( f(x) \) is all \( x eq \frac{1}{3} \).
• The range is all real numbers, as \( x \) approaches infinite values, so does \( y \).

Finding the inverse \( f^{-1}(x)=\frac{x}{1+3x} \) switches these roles. Now the domain is all real numbers except \( -\frac{1}{3} \), and the range returns to all \( y eq \frac{1}{3} \). Graphically this is evident in the asymptotes, lines the function can approach but never reach.

Algebraically verifying a function and its inverse involves checking that applying both one after the other returns you to the original value. Essentially, you compute both \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \).For the function \( f(x) = \frac{x}{1-3x} \) and its inverse \( f^{-1}(x) = \frac{x}{1+3x} \), you should verify by substitution.

• Calculate \( f(f^{-1}(x)) = \frac{\frac{x}{1+3x}}{1-3\left(\frac{x}{1+3x}\right)} \), simplify, and verify the result is \( x \).
• Conversely, compute \( f^{-1}(f(x)) = \frac{\frac{x}{1-3x}}{1+3\left(\frac{x}{1-3x}\right)} \), and confirm it simplifies to \( x \).

This algebraic process reassures that both functions are correctly stated, and they are proper inverses of each other. Not only does this reconfirm the solution but it strengthens your understanding of inverse relationships in functions.