Problem 50 - Algebra for College Students

Question

(a) find \(f^{-1}\) and (b) verify that \(\left(f \circ f^{-1}\right)(x)=x\) and \(\left(f^{-1} \circ f\right)(x)=x\). $$ f(x)=\frac{x}{x+1} \quad \text { for } x>-1 $$

Step-by-Step Solution

Answer

(a) \( f^{-1}(x) = \frac{-x}{x-1} \), (b) Both compositions equal x.

Step 1: Finding the inverse function definition

To find the inverse function, we start by writing the equation for the function and swap the roles of the independent variable (\(y\)) and dependent variable (\(x\)). Begin with: \( y = \frac{x}{x+1} \). Now, express \(x\) in terms of \(y\).

Step 2: Solve for x in terms of y

Rearrange the equation \(y = \frac{x}{x+1}\) to solve for \(x\). Multiply both sides by \(x+1\) to get \(y(x+1) = x\). This simplifies to \(yx + y = x\).

Step 3: Isolate x

Move all terms involving \(x\) to one side: \(yx - x = -y\). Factor out \(x\): \(x(y - 1) = -y\).

Step 4: Find x in terms of y

Divide both sides by \((y - 1)\) to solve for \(x\): \(x = \frac{-y}{y-1}\). This expression represents \(f^{-1}(y)\), or rewriting in terms of \(x\): \(f^{-1}(x) = \frac{-x}{x-1}\).

Step 5: Verify \((f \circ f^{-1})(x) = x\)

Compute \(f(f^{-1}(x))\): Substitute \(f^{-1}(x) = \frac{-x}{x-1}\) into \(f(x)\). Set \(x = \frac{-x}{x-1}\) in \(f(x) = \frac{x}{x+1}\) which simplifies as: \(f\left(\frac{-x}{x-1}\right) = \frac{\left(\frac{-x}{x-1}\right)}{\left(\frac{-x}{x-1}+1\right)}\). Simplify the inner expression.

Step 6: Simplify \(f(f^{-1}(x))\) to check it equals x

Simplify \(\left(\frac{-x}{x-1}+1\right)\) which becomes \(\left(\frac{-x + x - 1}{x-1}\right) = \left(\frac{-1}{x-1}\right)\). Thus, \(f\left(\frac{-x}{x-1}\right) = \frac{\left(\frac{-x}{x-1}\right)}{\left(\frac{-1}{x-1}\right)}\). Simplifying results in \(x\), so \((f \circ f^{-1})(x) = x\).

Step 7: Verify \((f^{-1} \circ f)(x) = x\)

Compute \(f^{-1}(f(x))\): Substitute \(f(x) = \frac{x}{x+1}\) into \(f^{-1}(x)\). \(f^{-1}(\frac{x}{x+1}) = \frac{-\left(\frac{x}{x+1}\right)}{\left(\frac{x}{x+1} - 1\right)}\). Simplify the inner expression.

Step 8: Simplify \(f^{-1}(f(x))\) to check it equals x

Simplify \(\frac{x}{x+1} - 1\), which becomes \(\frac{x - (x+1)}{x+1} = \frac{-1}{x+1}\). Thus, \(f^{-1}(\frac{x}{x+1}) = \frac{-\left(\frac{x}{x+1}\right)}{\left(\frac{-1}{x+1}\right)}\). After simplifying, this results in \(x\), so \((f^{-1} \circ f)(x) = x\).

Key Concepts

Function Composition

Function composition is a critical concept when working with inverse functions. In mathematics, composing functions involves applying one function to the results of another. It allows us to determine how the interactions of two functions affect an input.
For example, if you have two functions, like \(f(x)\) and \(g(x)\), then the composition \((f \circ g)(x)\) means that you first apply \(g\) to \(x\), and then \(f\) to the result of \(g(x)\). This is written as:
\[ (f \circ g)(x) = f(g(x)) \]
When dealing with inverse functions, this composition becomes particularly meaningful. If \(f(x)\) and \(f^{-1}(x)\) are inverse functions, composing them in either order should yield the original input \(x\). This is because an inverse function "undoes" the operation of the original function.
For instance:

• \((f \circ f^{-1})(x) = x\)
• \((f^{-1} \circ f)(x) = x\)

Understanding and verifying this property is essential because it confirms that you have correctly determined the inverse of a function.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations and expressions to solve for a desired variable. It is a fundamental skill in finding inverse functions.
Initially, you express your function equation in the form \(y = f(x)\). To find the inverse, you swap \(x\) and \(y\), and then solve for \(y\). The process of finding the inverse of the function \(f(x) = \frac{x}{x+1}\) illustrates this.

• Start by rewriting the function equation in terms of \(y\): \(y = \frac{x}{x+1}\).
• Swap \(x\) and \(y\) resulting in \(x = \frac{y}{y+1}\).
• Use algebraic manipulation to isolate \(y\). Multiply both sides by \(y+1\) to eliminate the denominator: \(x(y+1) = y\).
• Rearrange the equation: \(xy + x = y\).
• Factor terms involving \(y\): \(xy - y = -x\).
• Solve for \(y\): \(y(x-1) = -x\).
• Finally, divide by \((x-1)\) to isolate \(y\): \(y = \frac{-x}{x-1}\).

This resulting formula is the inverse function, \(f^{-1}(x) = \frac{-x}{x-1}\). Understanding these steps helps in conducting correct algebraic manipulation.

Verification of Inverse

Verification of an inverse function means confirming that your derived inverse truly "undoes" the original function. This verification is crucial as it ensures the function's and its inverse's correctness.
To verify, you need to check two conditions:

• \((f \circ f^{-1})(x) = x\)
• \((f^{-1} \circ f)(x) = x\)

Let’s consider \(f(x) = \frac{x}{x+1}\) and its inverse \(f^{-1}(x) = \frac{-x}{x-1}\). To verify, start with checking \((f \circ f^{-1})(x) = x\):

• Substitute \(f^{-1}(x)\) into \(f\), yielding \(f\left(\frac{-x}{x-1}\right)\).
• Simplify: \(\frac{\left(\frac{-x}{x-1}\right)}{1 + \frac{-x}{x-1}} = x\).

Next, check \((f^{-1} \circ f)(x) = x\):

• Substitute \(f(x)\) into \(f^{-1}\), resulting in \(f^{-1}\left(\frac{x}{x+1}\right)\).
• Simplify: \(\frac{-\left(\frac{x}{x+1}\right)}{1 - \frac{x}{x+1}} = x\).

When both conditions are satisfied, it confirms that the functions are indeed inverses of each other, demonstrating that the original function and its inverse properly cancel each other out, bringing us back to the initial input.