Problem 96 - Precalculus Essentials

Question

Show that $$f(x)=\frac{3 x-2}{5 x-3}$$ is its own inverse.

Step-by-Step Solution

Answer

For all real values of x, except \(x = \frac{3}{5}\), the function \(f(x) = \frac{3x - 2}{5x - 3}\) is its own inverse.

Step 1: Define the Function

First, let's define the given function, \(f(x) = \frac{3x - 2}{5x - 3}\).

Step 2: Compute f(f(x))

To prove that f(x) is its own inverse, we need to substitute \(f(x)\) again in place of x (i.e., we compute \(f(f(x))\)). So, we replace each x in the equation of \(f(x)\) by \(\frac{3x - 2}{5x - 3}\), which leads to: \(f(f(x)) = \frac{3(\frac{3x - 2}{5x - 3}) - 2}{5(\frac{3x - 2}{5x - 3}) - 3}\).

Step 3: Simplify f(f(x))

Next, we simplify the above equation: \(f(f(x)) = \frac{3(3x - 2) - 2(5x - 3)}{5(3x - 2) - 3(5x - 3)} = \frac{9x - 6 - 10x + 6}{15x - 10 - 15x + 9} = \frac{-x}{0}\). But we cannot divide by zero. So, we need a condition for the denominator not to be zero. Hence, 5x - 3 ≠ 0, which implies x ≠ \(\frac{3}{5}\).

Step 4: Find Solution

Finally, the solution will be x, for all real values of x except \(\frac{3}{5}\). In such case, function \(f(x)\) is termed as its own inverse.

Key Concepts

Composite Functions

A composite function is created when one function is applied to the result of another function. This means you are essentially combining two functions in a sequence. Think of it like a machine that does two tasks one after the other. If you have a function \( g(x) \) and another function \( f(x) \), the composite function is represented as \( (f \circ g)(x) = f(g(x)) \). It suggests that the function \( g(x) \) is evaluated first and then \( f(x) \) is applied to the result.

With the given problem, you went through this exact process when calculating \( f(f(x)) \). Here, you applied the function \( f(x) = \frac{3x - 2}{5x - 3} \) to itself, replacing the variable \( x \) with \( f(x) \). This created a new expression that represents the composite function:

• First, substitute every \( x \) in \( f(x) \) with \( \frac{3x - 2}{5x - 3} \), producing \( f(f(x)) = f\left(\frac{3x - 2}{5x - 3}\right) \).
• Next, simplify the result to see if it becomes the original input, \( x \), to verify if \( f(x) \) is indeed its own inverse.

Paying attention to how composite functions work is critical in understanding when one function can inverse the effect of another, especially when both functions are the same.

Rational Functions

Rational functions are functions that consist of ratios of two polynomials. In simpler terms, they are fractions where the numerator and the denominator are both polynomial expressions. A typical example was given in this problem, \( f(x) = \frac{3x - 2}{5x - 3} \), where both the numerator, \( 3x - 2 \), and the denominator, \( 5x - 3 \), are linear polynomials.

When working with rational functions, it's essential to be cautious of any restrictions on the variable \( x \), particularly values that cause division by zero. In this example, we saw that the function \( f(x) \) becomes undefined when the denominator is zero, meaning \( 5x - 3 eq 0 \). Solving this inequality demonstrates that \( x eq \frac{3}{5} \).

• This exclusion of \( x = \frac{3}{5} \) prevents undefined operations, which is important to remember whenever you deal with rational functions.
• The behavior and limitations of rational functions are key in tasks involving function inverses and solving composite functions.

Always keep the domain and any restrictions in mind when analyzing rational functions, as they directly impact the function's validity and the operations you can perform on them.

Function Inverses

A function inverse essentially "undoes" what the original function does. If a function \( f(x) \) maps input \( x \) to an output \( y \), then an inverse function \( f^{-1}(x) \) would map \( y \) back to \( x \). For a function to be its own inverse, applying it twice should return you to the initial value.

In this exercise, you needed to prove that \( f(x) = \frac{3x - 2}{5x - 3} \) is its own inverse by showing \( f(f(x)) = x \) for all permissible \( x \). This means if you replace \( x \) in \( f(x) \) with the full expression \( f(x) \) yields the original input \( x \).

• This involves calculating and simplifying \( f(f(x)) \) step by step, ensuring you carefully manage the algebraic manipulation involved.
• In this specific case, excluding the value \( x = \frac{3}{5} \), the composition \( f(f(x)) \) simplifies perfectly back to \( x \), proving that \( f(x) \) is indeed its own inverse.

An inverse function, therefore, is a powerful concept since it allows for reverse operations, undoing the effect of the function and assisting in solving equations involving functions.