Problem 15 - College Algebra Essentials

Question

The functions in Exercises \(11-28\) are all one-to-one. For each function, a. Find an equation for \(f^{-1}(x),\) the inverse function. b. Verify that your equation is correct by showing that \(f\left(f^{-1}(x)\right)=x\) and \(f^{-1}(f(x))=x\) $$f(x)=2 x+3$$

Step-by-Step Solution

Answer

The inverse function of \(f(x)=2x+3\) is \(f^{-1}(x)=\frac{x-3}{2}\). The verification process confirmed that \(f(f^{-1}(x))=x\) and \(f^{-1}(f(x))=x\).

Step 1: Compute Inverse Function

To find the inverse of \(f(x)=2x+3\), first replace \(f(x)\) with \(y\). Then, interchange the roles of \(x\) and \(y\), meaning replace \(y\) with \(x\) and vice versa. This leads to the equation \(x=2y+3\). Now solve this equation for \(y\) to find \(f^{-1}(x)\). Subtract 3 from both sides to get \(x-3=2y\). Then divide both sides by 2 to get \(y= \frac{x-3}{2}\). Hence, the inverse function is \(f^{-1}(x)=\frac{x-3}{2}\).

Step 2: Verify the inverse

To verify that \(f^{-1}(x)\) is indeed the inverse of \(f(x)\), check if \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\). For the first part, replace \(x\) in \(f(x)\) with \(f^{-1}(x)\) to get \(f(f^{-1}(x))=2f^{-1}(x)+3=2*\frac{x-3}{2}+3=x\). For the second part, replace \(x\) in \(f^{-1}(x)\) with \(f(x)\) to get \(f^{-1}(f(x))=\frac{f(x)-3}{2}=\frac{2x+3-3}{2}=x\). Therefore, \(f(f^{-1}(x))=x\) and \(f^{-1}(f(x))=x\) hold, which means \(f^{-1}(x)=\frac{x-3}{2}\) is indeed the inverse function of \(f(x)=2x+3\).

Key Concepts

One-to-One Functions

A crucial concept in understanding inverse functions is identifying one-to-one functions. A function is termed "one-to-one" if every unique input into the function gives a unique output. This implies that each output also maps back to one unique input. Such functions are essential when finding inverses because they assure us that each element in the range corresponds to exactly one element in the domain.

There are a few ways to determine if a function is one-to-one:

• Horizontal Line Test: If no horizontal line intersects the graph of the function more than once, it is one-to-one.
• Algebraic Check: Solve for whether the function means that if two outputs are equal, then the inputs must be equal (i.e., if \(f(a) = f(b)\) implies \(a = b\)).

Understanding this concept is fundamental before diving into finding inverse functions, as it ensures that each value has a unique reciprocal match.

Verification of Inverse Functions

Once you've calculated what seems to be the inverse function, verification is a necessary step to confirm your work. To verify the inverse function, two conditions must be satisfied:

• First, check whether \(f(f^{-1}(x)) = x\).
• Second, ensure that \(f^{-1}(f(x)) = x\).

This verification substantiates that applying the inverse function followed by the original function (or vice versa) returns the initial input. In our example with the function \(f(x) = 2x + 3\), after finding \(f^{-1}(x)\), we proved correctness by substitution and simplification:

• Replacing \(x\) in \(f(x)\) with \(f^{-1}(x)\) confirmed that the outcome was \(x\).
• Similarly, replacing \(x\) in \(f^{-1}(x)\) with \(f(x)\) confirmed the same outcome.

This dual check not only ensures mathematical accuracy but also aids in reinforcing the concept for better understanding.

Algebraic Manipulation

Algebraic manipulation forms the backbone of finding and verifying inverse functions. It involves rearranging equations and performing operations like addition, subtraction, multiplication, and division to isolate variables. In the context of finding inverse functions, the manipulation typically involves interchanging x and y in the original function's equation and then solving for y.

Let's see how it's done: Suppose you have \(y = 2x + 3\). Switch x and y to get \(x = 2y + 3\). The next step involves isolating y, which entails:

• Subtracting 3 from both sides: \(x - 3 = 2y\).
• Dividing both sides by 2: \(y = \frac{x - 3}{2}\).

Such manipulation is essential because it transforms the equation into a form that explicitly represents the inverse function. Being proficient in these algebraic techniques is vital not just for inverse functions, but is a fundamental skill in all algebraic problem solving.

College Algebra

College Algebra is a key element of a mathematics curriculum that often involves studies of more advanced functions, equations, and their properties. It serves as a bridge between basic algebra learned in high school and more diverse and complex topics in higher mathematics like calculus. Within college algebra, learners delve into subjects such as polynomial, rational, exponential, and logarithmic functions. The concept of inverse functions falls within this purview as students explore how functions can be reversed and applied to real-world contexts.

Topics in college algebra not only build on previous knowledge but also pave the way to a deeper understanding of mathematics. Learning to find and verify inverse functions is a good example of this as it involves applying previous algebraic skills in new ways. These exercises typically bolster a student's logical reasoning and problem-solving abilities, crucial for tackling higher-level math challenges and applications in science and engineering.