Problem 19 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Let \(f(x)=\frac{1}{x}\). Solve the following. (a) \(f\left(x^{2}\right)=1\) (b) \(-f(x)=f(x-1)\) (c) \(2 f(x-2)=f(x+3)\)

Step-by-Step Solution

Answer

So, the solutions for the given equations are: (a) \(x=\pm1\), (b) \(x= -\frac{1}{2}\), and (c) \(x = -5\).

Step 1: Solving equation (a)

Substitute \(f(x^{2})\) with \(\frac{1}{x^{2}}\). So, the equation \(f\left(x^{2}\right)=1\) turns to \(\frac{1}{x^{2}}=1\). Solving this equation gives \(x=\pm1\).

Step 2: Solving equation (b)

For equation (b) \(-f(x)=f(x-1)\), replace \(f(x)\) and \(f(x-1)\) with \(\frac{-1}{x}\) and \(\frac{1}{x-1}\) respectively. Solving the equation \(\frac{-1}{x}=\frac{1}{x-1}\), we get \(x = \frac{-1}{2}\).

Step 3: Solving equation (c)

For equation (c) \(2 f(x-2)=f(x+3)\), substitute \(f(x-2)\) and \(f(x+3)\) with \(\frac{1}{x-2}\) and \(\frac{1}{x+3}\) respectively. Solving this equation \(\frac{2}{x-2}=\frac{1}{x+3}\) yields \(x = \pm5\). Since \(x-2\) must not be zero as that would give us division by zero in the definition of our function, therefore we discard the solution \(x = 2\). Thus, the only valid solution here is \(x = -5\).

Key Concepts

Inverse Functions

Understanding inverse functions is essential when dealing with rational equations like those presented in the exercise. An inverse function essentially reverses the effect of the original function. For the function given in the exercise, \(f(x)=\frac{1}{x}\), the inverse function would do the opposite operation in order to return to the original input value.

This means if we apply the inverse function to \(f(x)\), we should get our input \(x\) back. Symbolically, if \(f^{-1}(x)\) is the inverse of \(f(x)\), then \(f^{-1}(f(x)) = x\). For the function \(f(x)=\frac{1}{x}\), its own inverse is itself, since \(\frac{1}{(\frac{1}{x})} = x\). This property is used implicitly when solving equations like \(f\left(x^{2}\right)=1\), where we need to 'undo' the function applied to the input \(x^{2}\).

One important aspect when working with inverse functions is to make sure the initial function is one-to-one, which means every output from the function corresponds to one and only one input. Only functions that are one-to-one have inverses that are also functions.

Function Composition

Function composition is another crucial concept that comes into play in our exercise. Composition of functions involves applying one function to the results of another function. We denote the composition of two functions \(f\) and \(g\) as \(f(g(x))\), which means 'first apply \(g\) to \(x\), then apply \(f\) to the result of \(g(x)\)'.

In the exercise, when we are asked to solve \(f\left(x^{2}\right)=1\), we are really dealing with a composition: the function \(f\) is applied to another function that squares the input, \(x^{2}\). To solve this, we tackled it as two separate operations: first the squaring of \(x\), followed by finding the value of \(f\) at that squared value.

Understanding how to apply functions in sequence, as in function compositions, is beneficial when trying to isolate the variable in complex equations or when dealing with nested functions in calculus problems.

Calculus Problems

The exercise additionally touches the edges of calculus problems, even though it's primarily focused on algebraic manipulation of rational functions. In calculus, you often work with more complex functions, but the principles remain the same — understanding how to manipulate and solve equations is foundational.

Many problems in calculus involve finding the derivatives or integrals of rational functions. For instance, if asked to find the derivative of our function \(f(x)=\frac{1}{x}\), one would apply the quotient rule. Similarly, if finding the integral (antiderivative) is the task, one might use partial fraction decomposition on more complex rational functions than the one provided in our exercise.

While this exercise does not require the use of calculus per se, the algebraic skills honed here are essential in tackling calculus problems involving derivatives and integrals of functions. Understanding how to solve for 'x' in rational equations builds a strong foundation for the functional manipulation frequently seen in calculus.