Problem 43 - 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 \(g^{\prime}(b)\) $$ f(x)=\left(x^{3}+1\right)^{3} ; \quad(1,8) $$

Step-by-Step Solution

Answer

In this exercise, we are given the function \(f(x) = \left(x^3 +1\right)^3\) and asked to verify that the point (1, 8) lies on its graph and find the derivative of its inverse function \(g'(b)\). First, to check if the point (1, 8) lies on the graph of the function \(f\), we plug in x = 1: \(f(1) = (1^3 + 1)^3 = 2^3 = 8\), which confirms that the point (1, 8) lies on the graph of the function \(f\). To find the derivative of \(f\), we use the chain rule: \(\frac{df}{dx} = 9x^2\left(x^3 + 1\right)^2\). Next, we evaluate the derivative at x = 1: \(\frac{df}{dx}(1) = 9(1)\left(2\right)^2 = 36\). Finally, we find the derivative of the inverse function \(g'(b)\) using the formula \(g'(b) = \frac{1}{f'(g(b))}\). Since g(8) = 1 and f'(1) = 36, we find that \(g'(8) = \frac{1}{f'(1)} = \frac{1}{36}\). Therefore, the derivative of the inverse function g'(b) is \(g'(8) = \frac{1}{36}\).

Step 1: Verify if the point (1, 8) lies on the graph of the function f(x)

Plug in x = 1 into the function \(f(x)\) and check if it results in an output of 8. $$ f(1) = \left((1)^3 + 1\right)^3 $$ Evaluating this yields: $$ f(1) = \left(1 + 1\right)^3 = 2^3 = 8 $$ Since \(f(1) = 8\), the point (1, 8) lies on the graph of the function \(f\).

Step 2: Find the derivative of the function f(x)

Differentiate the function \(f(x)\) with respect to x: $$ f(x) = \left(x^3 + 1\right)^3 $$ Apply the chain rule: $$ \frac{df}{dx} = 3\left(x^3 + 1\right)^2 \cdot 3x^2 $$ Simplify: $$ \frac{df}{dx} = 9x^2\left(x^3 + 1\right)^2 $$

Step 3: Evaluate the derivative at x = a = 1

Plug in x = 1 into the found derivative: $$ \frac{df}{dx}(1) = 9(1^2)\left((1)^3 + 1\right)^2 $$ Evaluating this yields: $$ \frac{df}{dx}(1) = 9(1)\left(2\right)^2 = 36 $$

Step 4: Find the derivative of the inverse function g'(b) using the formula

To find the derivative of the inverse function, use the following formula: $$ g'(b) = \frac{1}{f'(g(b))} $$ Since the point (1, 8) lies on the graph of the function \(f\), it means g(8) = 1. From Step 3, it is known that, f'(1) = 36. Use this information to calculate g'(b = 8): $$ g'(8) = \frac{1}{f'(g(8))} = \frac{1}{f'(1)} = \frac{1}{36} $$ So, the derivative of the inverse function g'(b) is \(g'(8) = \frac{1}{36}\).

Key Concepts

Derivatives

The concept of derivatives is a cornerstone in calculus, representing the rate of change or the slope of a function at a given point. When you see the notation \( f'(x) \), this means the derivative of \( f \) with respect to \( x \). Derivatives help us understand how a function behaves as it changes over its domain.

In the given exercise, we needed to find the derivative of the function \( f(x) = (x^3 + 1)^3 \). This required recognizing the function's structure and applying proper differentiation techniques. When differentiating complex expressions, it's crucial to use rules like the chain rule and product rule effectively.

Knowing how to calculate the derivative not only helps in finding rates of change but also aids in solving problems involving maxima and minima, as well as analyzing the concavity of graphs.

Chain Rule

The chain rule is an essential tool in calculus used to differentiate compositions of functions. If you have a function \( y = f(u) \) and another function \( u = g(x) \), the chain rule states that the derivative of \( y \) with respect to \( x \) is given by

• \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)

This formula helps in breaking down and simplifying the differentiation process.

In the original problem, we applied the chain rule to differentiate \( f(x) = (x^3 + 1)^3 \). By recognizing the outer function as \((u)^3\) and the inner function as \(u = x^3 + 1\), we effectively used the chain rule to obtain \( \frac{df}{dx} = 9x^2 (x^3 + 1)^2 \). This step was crucial in evaluating the derivative at specific points, allowing us to find the slope at \( x = 1 \) and contributing to determining the inverse function's derivative.

Calculus Concepts

Calculus is a field of mathematics that studies how things change and how to calculate these changes. It includes the study of derivatives, integrals, limits, and infinite series. Inverse functions and their derivatives are among the many intriguing concepts explored in calculus.

Inverse functions can reverse the effect of the original function. To find the derivative of an inverse function, the formula \( g'(b) = \frac{1}{f'(g(b))} \) is used. It's derived from the relationship that if a function \( f \) and its inverse \( g \) satisfy \( f(g(b)) = b \), then their respective derivatives relate inversely beyond just the function values.

Applying these concepts requires a deep understanding of how derivatives represent local linear approximations of functions and how these approximations behave when the input changes slightly. Mastering these principles opens doors to more advanced mathematical explorations.

Graph of a Function

Visualizing a function's behavior is often simplified by examining its graph. The graph of a function provides valuable insights into its properties, such as where the function increases or decreases, its maximum and minimum values, and points where it may be undefined.

In the problem, it was essential first to verify the point \((1, 8)\) on the graph of \( f(x) = (x^3 + 1)^3 \). This verification step confirmed that when we substitute \( x = 1 \) into the function, we indeed get \( y = 8 \). Understanding the graph also aids in visualizing how the derivative describes the function's slope and how steep or flat the curve is at specific points.

For the inverse function, picturing its graph involves reflecting the original function's graph over the line \( y=x \). This reflection helps illustrate how inputs and outputs of the function and its inverse navigate the Cartesian plane. Visual representations make the often abstract calculus concepts much more digestible.