Question

In Exercises 5-8, show that the slopes of the graphs of \(f\) and \(f^{-1}\) are reciprocals at the indicated points. $$\begin{array}{ll}\text { Function } & \text { Point } \\\\\hline f(x)=x^{3} & \left(\frac{1}{2}, \frac{1}{8}\right)\\\f^{-1}(x)=\sqrt[3]{x} & \left(\frac{1}{8}, \frac{1}{2}\right)\end{array}$$

Short answer

The slope of \(f(x)\) at the point \((1/2, 1/8)\) is 3/4, and the slope of \(f^{-1}(x)\) at the point \((1/8, 1/2)\) is 4/3. Since \(3/4 \times 4/3 = 1\), hence the slopes of the graphs of \(f\) and \(f^{-1}\) at the indicated points are reciprocals.

Step-by-step solution

Find the derivative

The first step is to find the derivative \(f'(x)\) of the function \(f(x) = x^3\). Using the power rule of derivatives, which states that the derivative of \(x^n\) is \(nx^{n-1}\), we get \(f'(x) = 3x^2\). Then plug in \(x = 1/2\) into \(f'(x)\) and solve to get the slope of \(f(x)\) at the point \((1/2, 1/8)\).

Find the derivative of the inverse function

The second step is to find the derivative of the inverse function \(f^{-1}(x) = \sqrt[3]{x}\) similarly. However, note that the derivative of \(\sqrt[3]{x}\) is \((1/3)x^{-2/3}\). Then plug in \(x = 1/8\) into \(f'^{-1}(x)\) to find the slope of \(f^{-1}(x)\) at the point \((1/8, 1/2)\).

Show that the slopes are reciprocals

In the final step, compare the slopes obtained from step 1 and step 2 to confirm that they are indeed reciprocals. Two numbers are reciprocals if their product equals 1. In other words, when one number is divided by the other, the result is 1. Show that when the slope of \(f(x)\) is divided by the slope of \(f^{-1}(x)\), or vice versa, the result is 1.

Key concepts

Derivatives of Inverse Functions

The concept of derivatives of inverse functions is vital in calculus. Inverse functions reverse the effect of the original function. Consider a function like f(x), which has an inverse denoted as f^-1(x). To find the derivative of this inverse function, there are few ways, but one common method is to use the relationship f'(f^-1(y))⋅f'^-1(y) = 1. This equation signifies that the derivative of the function and its inverse at corresponding points are multiplicative inverses (reciprocals).

In the exercise, we first find the derivative of the function f(x) = x³, which is a straightforward application of the power rule. Moreover, the derivative of the inverse function f^-1(x) = ∛x can be found using the chain rule in conjunction with the knowledge of the original function's derivative. Such practices reinforce the understanding of the intricate relationship between a function and its inverse when it comes to their derivatives.

Reciprocal Relationship of Slopes

The reciprocal relationship of slopes between a function and its inverse is a fascinating concept. If we have a function f and its inverse f^-1, at any corresponding points on their respective graphs, the slopes of the tangent lines at those points are reciprocal to each other.

This can be visualized by plotting both functions on a graph and examining the slopes at the points that reflect each other across the line y = x. This is because inverse functions are symmetric with respect to this line. In our example, the slope of the original function f(x) at point (1/2, 1/8) and the slope of f^-1(x) at point (1/8, 1/2) are shown to be reciprocals. By multiplying these slopes, we confirm that their product is indeed 1, showcasing this reciprocal property which is foundational in understanding the behavior of inverse functions in calculus.

Power Rule of Derivatives

An important rule in differentiation is the power rule. This rule states that for any function in the form of f(x) = xⁿ, where n is any real number, the derivative f'(x) is given by f'(x) = nx^n-1. This rule is essential when taking derivatives of functions with variable exponents and is one of the first rules taught in calculus due to its straightforward application and utility.

In our provided exercise, we use the power rule to differentiate the function f(x) = x³. According to the power rule, we bring down the exponent as a coefficient and decrease the power by one, resulting in f'(x) = 3x². This is an elegant simplification for finding slopes of functions at specific points and is a gateway to solving more complex problems that involve rates of change and graphical analysis of functions.