Question

\(f(x)=\frac{1}{x} ; g(x)=\left(x+\frac{1}{x}\right)\), then which of the following is true? (a) \(f(g(x)) \cdot g(f(x))=1\) (b) \(\frac{f(g(x))}{g(f(x))}=1\) (c) \(f(g(x))+g(f(x))=1\) (d) none of these

Short answer

a) f(g(x)) ⋅ g(f(x)) = 1 b) \(\frac{f(g(x))}{g(f(x))} = 1\) c) f(g(x)) + g(f(x)) = 1 Answer: b) \(\frac{f(g(x))}{g(f(x))} = 1\)

Step-by-step solution

Calculate f(g(x))

To find the composition f(g(x)), we substitute g(x) into the function f(x). So, f(g(x)) = f(x + 1/x). Using the given formula for f(x) = 1/x, we can rewrite f(g(x)) as: f(g(x)) = \(\frac{1}{x + \frac{1}{x}}\)

Calculate g(f(x))

To find the composition g(f(x)), we substitute f(x) into the function g(x). So, g(f(x)) = g(1/x). Using the given formula for g(x) = x + 1/x, we can rewrite g(f(x)) as: g(f(x)) = \(\left(\frac{1}{x}+\frac{1}{\frac{1}{x}}\right)\) = \(\frac{1}{x}+x\)

Verify the statements

Now, we will use the compositions f(g(x)) and g(f(x)) to verify which statement is true. (a) Let's check if f(g(x)) ⋅ g(f(x)) = 1. \(f(g(x)) \cdot g(f(x)) = \left(\frac{1}{x + \frac{1}{x}}\right) \cdot \left(\frac{1}{x}+x\right) = \frac{\left(\frac{1}{x}+x\right)}{\left(x^2 +1\right)}\) As this expression does not simplify to 1, statement (a) is false. (b) Let's check if \(\frac{f(g(x))}{g(f(x))} = 1\). \(\frac{\frac{1}{x + \frac{1}{x}}}{\frac{1}{x} + x} = \frac{1}{\left(x^2 + 1\right)} \div \frac{x^2 + 1}{\left(x^2 + 1\right)} = \frac{1}{x^2 + 1} \cdot \frac{\left(x^2 + 1\right)}{x^2 + 1}\) This expression simplifies to 1, so statement (b) is true. (c) Let's check if f(g(x)) + g(f(x)) = 1. \(f(g(x)) + g(f(x)) = \left(\frac{1}{x + \frac{1}{x}}\right) + \left(\frac{1}{x}+x\right) = \frac{1}{x^2 + 1} + \left(\frac{1}{x}+x\right)\) This expression does not simplify to 1, so statement (c) is false. As statement (b) is true, the correct answer is (b) \(\frac{f(g(x))}{g(f(x))}=1\).

Key concepts

Function Composition

Understanding the concept of function composition is essential in mathematics, as it allows us to combine two or more functions to form a new function. When we compose two functions, say f and g, the result is a new function where we apply g first and then apply f to the result of g. This is notated as f(g(x)), read as 'f of g of x'.

It's crucial to follow the order of operations carefully, just like following a recipe step by step. If you swap the functions, you'll get a different result, just as if you mixed up the steps in a recipe. For instance, g(f(x)) does not generally equal f(g(x)). In our textbook exercise, we saw an application of function composition and how to evaluate composite functions methodically.

Quantitative Aptitude

Quantitative aptitude is all about your ability to tackle numerical and logical problems. It plays a pivotal role not just in academic assessments but also in various competitive exams and day-to-day problem solving. Being able to work through composite functions, as shown in our example, is a great display of quantitative aptitude.

This skill involves algebraic reasoning and an understanding of how different mathematical concepts interact with each other. Enhancing your quantitative aptitude means improving your critical thinking skills and becoming more adept at mathematical reasoning. Encountering a problem like the one from our textbook exercises provides a perfect opportunity to refine these skills.

Mathematical Functions

The world of mathematical functions is vast and varied, encompassing everything from simple linear equations to complex polynomial expressions. Functions represent relationships between sets of numbers or variables. They're fundamental building blocks in various fields, including engineering, economics, and physics.

Our textbook problem dealt with reciprocal and sum functions. By understanding different types of functions and their behaviors, we can solve a wide range of problems efficiently. The reciprocal function, f(x)=1/x, for example, has a profound importance in mathematics due to its unique properties and its role in defining the very concept of division as the inversion of multiplication.

As you progress in mathematics, you'll encounter functions that can model real-world scenarios, from the growth of investments to the physics of motion. Building a strong foundation in understanding and manipulating functions is key to your success in the field.