Problem 39 - Single Variable Calculus

Question

Let \(g\) denote the inverse of the function \(f\). (a) Show that the point \((a, b)\) lies on the graph of \(f .\) (b) Find \(g^{\prime}(b)\) $$ f(x)=2 x+1 ; \quad(2,5) $$

Step-by-Step Solution

Answer

The point \((2, 5)\) lies on the graph of the function \(f(x) = 2x + 1\) because \(f(2) = 5\). The inverse function is \(g(x) = \frac{x - 1}{2}\) and its derivative, \(g'(x) = \frac{1}{2}\). Therefore, \(g'(b) = g'(5) = \frac{1}{2}\).

Step 1: Verify the Point on the Graph of the Function

To show that the point \((2, 5)\) lies on the graph of the function \(f(x) = 2x + 1\), we will substitute the \(x\) and \(y\) coordinates of this point into the function and verify that it holds: If \((a, b) = (2, 5)\), then \(f(a) = f(2) = 2(2) + 1 = 5\). Since \(f(a) = b\), the point \((2, 5)\) lies on the graph of the function \(f\).

Step 2: Find the Inverse Function, g(x)

To find the inverse function, \(g(x)\), we will solve the equation \(y = f(x) = 2x + 1\) for \(x\). 1. Replace \(f(x)\) with \(y\): \(y = 2x + 1\). 2. Interchange \(x\) and \(y\): \(x = 2y + 1\). 3. Solve for \(y\): \(y = \frac{x - 1}{2}\). Thus, the inverse function, \(g(x)\), is \(g(x) = \frac{x - 1}{2}\).

Step 3: Calculate the Derivative of g(x)

Next, we will find the derivative of \(g(x)\) with respect to \(x\): $$ g'(x) = \frac{d}{dx} \left(\frac{x-1}{2}\right) $$ Here, we have a constant in the denominator, so we will treat the entire expression as a single constant multiplied by \(x - 1\): $$ g'(x) = \frac{1}{2} \frac{d}{dx} (x-1) $$ By applying the Chain Rule, we find: $$ g'(x) = \frac{1}{2} (1) $$ So, \(g'(x) = \frac{1}{2}\).

Step 4: Find g'(b)

Since we have now found the derivative of \(g(x)\), we can substitute the value of \(b\) (5) into \(g'(x)\): $$ g'(b) = g'(5) = \frac{1}{2} $$ Hence, the derivative of the inverse function, \(g'(b)\), is \(\frac{1}{2}\).

Key Concepts

Derivatives

Derivatives are fundamental in calculus. They measure how a function changes as its input changes. Imagine it as the function's response rate. When we say "the derivative of a function," we're referring to a new function that tells us how our original function behaves at every point on its graph. For example, when finding the derivative of the inverse function \(g(x) = \frac{x - 1}{2}\), it becomes a linear transformation: \(g'(x) = \frac{1}{2}\). This tells us that for any change in \(x\), \(g(x)\) changes at half that rate. This simple yet powerful tool helps us understand, predict, and "see" how different functions behave. The derivative of our inverse function is key to answering specific questions, such as how the inverse function's output responds to changes in its input.

Function Graphs

Function graphs offer a pictorial view of an equation’s behavior. Graphing a function like \(f(x) = 2x + 1\) gives you a straight line because it is a linear function. The slope of the line is 2, which means for every unit increase in \(x\), \(f(x)\) increases by 2 units. When checking points on the graph, like \((2, 5)\), substitute \(x=2\) into \(f(x)\) and see if it gives \(y=5\). If it holds, as it does here, the point lies on the graph. This confirms the graph fits this equation. Graphs not only show us individual points but also help visualize how different functions relate, such as how \(f(x)\) and its inverse, \(g(x)\), might look in the coordinate plane.

Chain Rule

The Chain Rule is a rule for calculating the derivative of the composition of two or more functions. Think of it as a process for finding the rate of change in a scene with multiple actors. It helps when dealing with complex functions that can be broken down into simpler pieces. In the situation of the inverse function \(g(x)\), the expression is already quite simple. The derivative \(g'(x) = \frac{1}{2} (1)\) doesn't require complex applications of the Chain Rule, but it's worth noting that even straightforward derivatives utilize these principles. If you find scenarios where a function is nested within another, like \(g(h(x))\), the Chain Rule provides the method: take the derivative of the outer function, multiply by the derivative of the inner one. This powerful tool is versatile, simplifying how we approach derivatives in multi-layered functions.

Calculating Inverses

Finding an inverse function essentially reverses the process of a function. When we calculate an inverse, we're swapping inputs and outputs to create a function that "undoes" the work of the original. The process involves swapping \(x\) and \(y\) and then solving the resulting equation for \(y\). In our original function \(f(x) = 2x + 1\), its inverse is obtained by first writing \(y = 2x + 1\), then switching \(x\) and \(y\) to form \(x = 2y + 1\), and lastly solving for \(y\), giving us \(g(x) = \frac{x - 1}{2}\).Understanding inverses is crucial, as they provide insights into solving equations and offer solutions to real-world problems where processes need reversing or outcomes need reverting. The practice of calculating inverses is essential in algebra and calculus, providing a deeper grasp of function interactions.