Problem 36 - Algebra and Trigonometry

Question

Use the Inverse Function Property to show that \(f\) and \(g\) are inverses of each other. $$ f(x)=\frac{x-5}{3 x+4}, \quad g(x)=\frac{5+4 x}{1-3 x} $$

Step-by-Step Solution

Answer

f and g are inverses because f(g(x)) = x and g(f(x)) = x.

Step 1: Apply Function Composition

To show that two functions, \(f\) and \(g\), are inverses of each other, we must prove two conditions: \(f(g(x)) = x\) and \(g(f(x)) = x\). Let's start by composing \(f(g(x))\).

Step 2: Substitute g(x) into f(x)

Substitute \(g(x) = \frac{5 + 4x}{1 - 3x}\) into \(f(x)\):\[f(g(x)) = f\left(\frac{5 + 4x}{1 - 3x}\right) = \frac{\frac{5 + 4x}{1 - 3x} - 5}{3\left(\frac{5 + 4x}{1 - 3x}\right) + 4}\]

Step 3: Simplify the Numerator

Let's simplify the numerator:\[\frac{5 + 4x}{1 - 3x} - 5 = \frac{5 + 4x - 5(1 - 3x)}{1 - 3x} = \frac{5 + 4x - 5 + 15x}{1 - 3x} = \frac{19x}{1 - 3x}\]

Step 4: Simplify the Denominator

Now, simplify the denominator:\[3\left(\frac{5 + 4x}{1 - 3x}\right) + 4 = \frac{3(5 + 4x) + 4(1 - 3x)}{1 - 3x} = \frac{15 + 12x + 4 - 12x}{1 - 3x} = \frac{19}{1 - 3x}\]

Step 5: Compute f(g(x))

Now compute:\[f(g(x)) = \frac{19x}{1-3x} \div \frac{19}{1-3x} = \frac{19x}{1-3x} \times \frac{1-3x}{19} = x\]Thus, \(f(g(x)) = x\).

Step 6: Apply Composition in Reverse

Now let's prove \(g(f(x)) = x\) by substituting \(f(x)\) into \(g(x)\):\[g(f(x)) = g\left(\frac{x-5}{3x+4}\right) = \frac{5 + 4\left(\frac{x-5}{3x+4}\right)}{1 - 3\left(\frac{x-5}{3x+4}\right)}\]

Step 7: Simplify the Reverse Composition

Simplify both the numerator and denominator for \(g(f(x))\). For the numerator:\[5 + \frac{4(x-5)}{3x+4} = \frac{5(3x+4) + 4(x-5)}{3x+4} = \frac{15x + 20}{3x+4}\]For the denominator:\[1 - \frac{3(x-5)}{3x+4} = \frac{3x+4 - 3(x-5)}{3x+4} = \frac{19}{3x+4}\]

Step 8: Compute g(f(x))

Now compute:\[g(f(x)) = \frac{15x + 20}{3x+4} \div \frac{19}{3x+4} = \frac{15x + 20}{3x+4} \times \frac{3x+4}{19} = x\]Thus, \(g(f(x)) = x\).

Step 9: Conclusion

Since we have shown that \(f(g(x)) = x\) and \(g(f(x)) = x\), \(f\) and \(g\) are indeed inverses of each other.

Key Concepts

Function Composition

Function composition involves combining two functions in such a way that the output of one function becomes the input of another. To demonstrate that functions \(f\) and \(g\) are inverses, we need to evaluate \(f(g(x))\) and \(g(f(x))\). This means:

• First, plug the value of \(g(x)\) into \(f(x)\). This requires substituting \(g(x)\) for each \(x\) in the function \(f(x)\), essentially creating a new combined function.
• Second, reverse the process by substituting \(f(x)\) into \(g(x)\).

If after performing these operations, both \(f(g(x)) = x\) and \(g(f(x)) = x\), this means \(f\) and \(g\) are inverse functions. Successfully composing these functions and arriving at \(x\) indicates they undo each other's operations.

Algebraic Simplification

Algebraic simplification is essential in making complex expressions manageable and understandable. It involves systematically reducing an expression to its simplest form. In the context of inverse functions, simplifying the function composition \(f(g(x))\) or \(g(f(x))\) to equal \(x\) requires these steps:

• Combine like terms: This means gathering similar terms together, making the function easier to read and compute.
• Eliminate unnecessary terms: Remove any terms in the numerator or the denominator that add no value to reaching the simplest form.
• Simplify fractions: Convert complex fractions into a more straightforward expression by using common denominators and factoring when necessary.

These steps were employed in the exercise to show that operations within \(f(x)\) and \(g(x)\) cancel each other out effectively, confirming they are inverses by simplifying down to \(x\).

Numerator and Denominator

Dealing with the numerator and denominator in algebraic expressions is a fundamental part of function simplification. Each function can be expressed as a fraction where the numerator and denominator must be handled separately before simplification:

• Numerator: This is the top part of the fraction. Focus on combining like terms, as was done with \(5 + 4x - 5\) in the numerator of \(f(g(x))\).
• Denominator: The bottom part of the fraction, simplify it similarly by combining and reducing terms, such as \(3\left(\frac{5 + 4x}{1 - 3x}\right) + 4\).
• Cross-multiplication: When solving expressions like \(f(g(x))\), multiply the numerator of one fraction by the denominator of the other and vice versa to clear out complex fractions.

By ensuring both the numerator and the denominator are simplified independently and then together, it becomes possible to solve expressions efficiently. This was crucial in validating that \(f(x)\) and \(g(x)\) are inverses, confirming the simple form \(x\) after composition.