Finding The Inverse Function Of F(x) = 7/(5+x) And Determining Domain And Range

In mathematics, understanding the concept of one-to-one functions is crucial for grasping the idea of inverse functions. A function is considered one-to-one, or injective, if each element in the range corresponds to exactly one element in the domain. In simpler terms, if f(x₁) = f(x₂), then x₁ must equal x₂. This property is essential because only one-to-one functions have inverses. The inverse function, denoted as f⁻¹(x), essentially reverses the operation of the original function. If f(a) = b, then f⁻¹(b) = a. Finding the inverse function involves swapping the roles of x and y in the original function's equation and then solving for y. This process reveals the function that undoes the operation performed by the original function.

When analyzing functions and their inverses, the domain and range play vital roles. The domain of a function is the set of all possible input values (x-values), while the range is the set of all possible output values (y-values). For an inverse function, the domain is the range of the original function, and the range is the domain of the original function. This reciprocal relationship highlights how the inverse function essentially mirrors the behavior of the original function. In this article, we will delve into the function f(x) = 7/(5+x), demonstrate that it is one-to-one, find its inverse function, and determine the domain and range of both the function and its inverse.

Part (a): Proving f(x) = 7/(5+x) is One-to-One and Finding its Inverse

To prove that f(x) = 7/(5+x) is a one-to-one function, we need to show that if f(x₁) = f(x₂), then x₁ = x₂. This involves a simple algebraic proof:

Assume f(x₁) = f(x₂).

Then, 7/(5+x₁) = 7/(5+x₂).

To solve this equation, we can cross-multiply:

7(5+x₂) = 7(5+x₁).

Divide both sides by 7:

5+x₂ = 5+x₁.

Subtract 5 from both sides:

x₂ = x₁.

Since x₁ = x₂, we have proven that f(x) = 7/(5+x) is indeed a one-to-one function. This crucial step allows us to proceed with finding the inverse function.

Now, to find the inverse function, f⁻¹(x), we follow these steps:

1. Replace f(x) with y: y = 7/(5+x).

2. Swap x and y: x = 7/(5+y).

3. Solve for y:

   • Multiply both sides by (5+y): x(5+y) = 7.
   • Distribute x: 5x + xy = 7.
   • Isolate the term with y: xy = 7 - 5x.
   • Divide by x: y = (7-5x)/x.

Therefore, the inverse function is f⁻¹(x) = (7-5x)/x. This inverse function undoes the operation of the original function, meaning if we input a value into f(x) and then input the result into f⁻¹(x), we should obtain our original input value.

To check our answer, we can verify that f(f⁻¹(x)) = x and f⁻¹(f(x)) = x. Let's start with f(f⁻¹(x)): f(f⁻¹(x)) = f((7-5x)/x) = 7/(5 + (7-5x)/x)

To simplify this expression, we need to find a common denominator in the denominator:

f(f⁻¹(x)) = 7/((5x + 7 - 5x)/x) = 7/(7/x) = 7 * (x/7) = x.

Next, we verify f⁻¹(f(x)) = x: f⁻¹(f(x)) = f⁻¹(7/(5+x)) = (7 - 5(7/(5+x)))/(7/(5+x)).

Again, we simplify this expression:

f⁻¹(f(x)) = (7(5+x) - 35)/(7) * ((5+x)/7) = (35 + 7x - 35)/(7/(5+x)) = (7x/(5+x)) * ((5+x)/7) = x.

Since both f(f⁻¹(x)) = x and f⁻¹(f(x)) = x, we have successfully verified that f⁻¹(x) = (7-5x)/x is indeed the inverse function of f(x) = 7/(5+x). This confirmation step is essential in ensuring the correctness of our derived inverse function.

Part (b): Determining the Domain and Range of f(x) and f⁻¹(x)

The domain of a function is the set of all possible input values (x-values) for which the function is defined. For f(x) = 7/(5+x), the function is undefined when the denominator is zero. Thus, we need to find the values of x for which 5+x = 0.

Solving for x, we get x = -5. Therefore, the domain of f(x) is all real numbers except -5. We can express this in interval notation as (-\infty, -5) ∪ (-5, \infty).

The range of a function is the set of all possible output values (y-values). To find the range of f(x), we consider the behavior of the function as x approaches different values. As x approaches -5 from the left, f(x) approaches negative infinity, and as x approaches -5 from the right, f(x) approaches positive infinity. As x approaches positive or negative infinity, f(x) approaches 0. However, f(x) never actually equals 0 because the numerator is a constant (7).

Therefore, the range of f(x) is all real numbers except 0, which can be written in interval notation as (-\infty, 0) ∪ (0, \infty).

Now, let's determine the domain and range of the inverse function, f⁻¹(x) = (7-5x)/x. The domain of f⁻¹(x) is the set of all x-values for which the function is defined. Similar to f(x), f⁻¹(x) is undefined when the denominator is zero. So, we need to find the values of x for which x = 0. Thus, the domain of f⁻¹(x) is all real numbers except 0, which is (-\infty, 0) ∪ (0, \infty).

To find the range of f⁻¹(x), we can use the relationship that the range of the inverse function is the domain of the original function. Therefore, the range of f⁻¹(x) is the same as the domain of f(x), which is (-\infty, -5) ∪ (-5, \infty). This reciprocal relationship between the domains and ranges of a function and its inverse is a fundamental concept in understanding inverse functions.

In summary:

• Domain of f(x): (-\infty, -5) ∪ (-5, \infty)
• Range of f(x): (-\infty, 0) ∪ (0, \infty)
• Domain of f⁻¹(x): (-\infty, 0) ∪ (0, \infty)
• Range of f⁻¹(x): (-\infty, -5) ∪ (-5, \infty)

These results demonstrate the interconnectedness of a function and its inverse. The domain of one is the range of the other, and vice versa. This relationship is crucial for a comprehensive understanding of functions and their inverses.

Conclusion: The Significance of Inverse Functions

In conclusion, we have successfully demonstrated that the function f(x) = 7/(5+x) is one-to-one and found its inverse function, f⁻¹(x) = (7-5x)/x. We also determined the domain and range of both functions, highlighting the reciprocal relationship between them. Understanding inverse functions is crucial in various areas of mathematics and its applications. Inverse functions allow us to reverse mathematical processes, which is essential in solving equations, understanding transformations, and many other contexts.

The concept of one-to-one functions is fundamental to the existence of inverse functions. If a function is not one-to-one, it does not have a unique inverse. This is because the inverse function would not be able to uniquely map the output back to the input. The process of finding an inverse function involves swapping the roles of the input and output variables and solving for the new output variable. This algebraic manipulation reveals the function that undoes the original function's operation.

The domain and range of a function and its inverse provide valuable information about the behavior of these functions. The domain represents the set of all permissible inputs, while the range represents the set of all possible outputs. For an inverse function, the domain and range are swapped compared to the original function. This reciprocal relationship underscores the nature of inverse functions as reversed mappings.

The ability to find and analyze inverse functions is a powerful tool in mathematical problem-solving. It allows us to tackle problems from different perspectives and gain deeper insights into the relationships between variables. In fields such as calculus, differential equations, and complex analysis, inverse functions play a crucial role in various theorems and techniques. Furthermore, the concept of inverse functions extends beyond mathematics and finds applications in areas such as computer science, engineering, and economics.

By mastering the concepts of one-to-one functions, inverse functions, domains, and ranges, we equip ourselves with essential tools for mathematical analysis and problem-solving. The example of f(x) = 7/(5+x) serves as a valuable illustration of these concepts and their applications. As we continue our mathematical journey, a thorough understanding of inverse functions will undoubtedly prove to be invaluable in tackling more complex and challenging problems.