Question

In Exercises \(13-24,\) find \(f^{-1}\) and verify that $$\left(f \circ f^{-1}\right)(x)=\left(f^{-1} \circ f\right)(x)=x$$ \(f(x)=2 x+3\)

Short answer

The inverse function is \(f^{-1}(x) = (x-3)/2\). The conditions \((f \circ f^{-1})(x)=x\) and \((f^{-1} \circ f)(x)=x\) are validated.

Step-by-step solution

Find the inverse function

To find the inverse of \(f(x) = 2x + 3\), we need to replace \(f(x)\) with \(y\). We'll get an equation \(y = 2x + 3\). To find the inverse, we need to swap the roles of \(x\) and \(y\). This will give us \(x = 2y + 3\). After that, we need to solve it for \(y\) which will be our \(f^{-1}(x)\). Solving for \(y\) we get: \(y=(x-3)/2\). So, the inverse function \(f^{-1}(x) = (x-3)/2\).

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

We need to verify \((f \circ f^{-1})(x) = x\). It means that we have to apply the function \(f\) to its inverse \(f^{-1}\). Given \(f^{-1}(x) = (x-3)/2\), we plug this into \(f(x)\). So, \((f \circ f^{-1})(x) = f(f^{-1}(x)) = f((x-3)/2)\). We replace \(f(x)\) by its formula: \(2x + 3\). Replacing \(x\) with \((x-3)/2\), we get \(2*((x-3)/2) + 3 = (x-3) + 3 = x\). So, \((f \circ f^{-1})(x) = x\) is verified.

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

In this step, we need to verify \((f^{-1} \circ f)(x) = x\). It implies that we have to apply the inverse function \(f^{-1}\) to \(f(x)\). Given \(f(x) = 2x + 3\), we put this in \(f^{-1}(x)\). So, \((f^{-1} \circ f)(x) = f^{-1}(f(x)) = f^{-1}(2x + 3)\). Now we replace \(f^{-1}(x)\) by its formula: \((x-3)/2\). Replacing \(x\) with \(2x + 3\), we get \((2x+3-3)/2 = 2x/2 = x\). Hence, \((f^{-1} \circ f)(x) = x\) is analyzed.

Key concepts

Function Composition

Function composition is a way of combining two functions in which the output of one function becomes the input of the other. For the uninitiated, it can seem like a magical process, but it's really just about connecting operations. In practice, when we have two functions, say f and g, we can form the composite function \( (f \circ g)(x) \) by applying g to x first and then applying f to the result. It's like a relay race in the world of functions, with the output of one becoming the baton handed to the next.

This idea is crucial when discussing inverse functions. In the exercise, we explored the composition of function f with its inverse f^-1 to verify the property \( (f \circ f^{-1})(x) = (f^{-1} \circ f)(x) = x \). When a function is composed with its inverse, the result should be the identity function, which essentially returns the original input. To make it stick, imagine doing a round trip: you go somewhere using one route (f) and come back via the inverse route (f^-1); you should end up exactly where you started!

Algebraic Manipulation

Algebraic manipulation is a suite of techniques for transforming equations and expressions. It's the bread and butter of solving for variables, rearranging formulas, and simplifying complex expressions. To effectively tackle algebra, one must be a maestro of these maneuvers: adding, subtracting, multiplying, dividing, and factoring, among others.

In the context of the given exercise, we used algebraic manipulation to find the inverse function of f(x). Starting with the linear function f(x) = 2x + 3, we followed a systematic process: we switched x and y and then solved for y, representing the inverted relationship. The manipulation here was straightforward, involving subtracting 3 from both sides and then dividing by 2 to isolate y, yielding the inverse function f^-1(x) = (x-3)/2. The ability to manipulate algebraic expressions is not just academic gymnastics; it's a critical skill that allows us to extract and isolate the pieces of information we really need.

Verifying Inverse Function

Verifying that two functions are inverses of each other is a bit like confirming a secret handshake. It's not enough that they look similar backward and forward; they must perfectly undo each other's actions. When verifying an inverse function, our goal is to show that when we compose the original function with its proposed inverse, we get the identity function—meaning we are back where we started, with x.

The exercise tasked us with proving this, and we showed that both \( (f \circ f^{-1})(x) \) and \( (f^{-1} \circ f)(x) \) return x. It's like a perfectly executed U-turn. For clarity, when we say \( f^{-1}(f(x)) = x \), we're saying that the inverse function should reverse whatever f did, no more and no less—effectively canceling it out. By confirming that both composites give us back our original x, we cement the fact that f^-1 is truly the inverse of f. Understanding this concept ensures that we don't just follow equations blindly but grasp the deeper interplay between functions and their inverses.