Question

Suppose that \(f\) is differentiable and that there are real numbers \(x_{1}\) and \(x_{2}\) such that \(f\left(x_{1}\right)=x_{2}\) and \(f\left(x_{2}\right)=x_{1}\). Let \(g(x)=f(f(f(f(x))))\). Show that \(g^{\prime}\left(x_{1}\right)=g^{\prime}\left(x_{2}\right)\).

Short answer

Since \(g'(x_1) = (f'(x_1) \cdot f'(x_2))^2\) and \(g'(x_2) = (f'(x_2) \cdot f'(x_1))^2\), we have \(g'(x_1) = g'(x_2)\).

Step-by-step solution

Define the function iteratively

We are given a function \(f\) such that \(f(x_1) = x_2\) and \(f(x_2) = x_1\). Define the function \(g(x) = f(f(f(f(x))))\). To solve the problem, we need to understand how \(g(x)\) behaves at \(x_1\) and \(x_2\). By substituting \(x_1\) and \(x_2\), we can find patterns or simplify calculations.

Substitute known points into g(x)

Substitute \(x_1\) into \(g(x)\): \(g(x_1) = f(f(f(f(x_1)))) = f(f(f(x_2))) = f(f(x_1)) = f(x_2) = x_1\). Similarly, for \(x_2\): \(g(x_2) = f(f(f(f(x_2)))) = f(f(f(x_1))) = f(f(x_2)) = f(x_1) = x_2\). Thus, \(g(x_1) = x_1\) and \(g(x_2) = x_2\).

Differentiate g(x) using the chain rule

Differentiate \(g(x) = f(f(f(f(x))))\) using the chain rule: \[g'(x) = f'(f(f(f(x)))) \cdot f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x)\]. This expression represents how the derivative of the outer function \(f\) is multiplied by the repeated inner applications of \(f\).

Evaluate g'(x) at specific points

Evaluate \(g'(x)\) at \(x_1\) and \(x_2\). Since \(f(x_1) = x_2\) and \(f(x_2) = x_1\), it follows that: \[g'(x_1) = f'(x_2) \cdot f'(x_1) \cdot f'(x_2) \cdot f'(x_1)\] and \[g'(x_2) = f'(x_1) \cdot f'(x_2) \cdot f'(x_1) \cdot f'(x_2)\].

Simplify and compare derivatives

Notice the symmetry in the derivatives: \[g'(x_1) = (f'(x_1) \cdot f'(x_2))^2\] and \[g'(x_2) = (f'(x_2) \cdot f'(x_1))^2\]. These are equal because multiplication is commutative, and it gives \(g'(x_1) = g'(x_2)\). Therefore, the derivatives are equal at these points.

Key concepts

Chain Rule

The chain rule is a fundamental technique used in calculus to find the derivative of composite functions. When we have a function composed of several nested functions, the chain rule helps us differentiate it step by step. In the given problem, we apply the chain rule to a function involving multiple layers of the function \( f \), specifically \( g(x) = f(f(f(f(x)))) \).

Here's how it works:

• Identify the "outer" and "inner" functions. Here, \( f \) is repeated multiple times as the inner function, while its last application serves as the outer layer in \( g(x) \).
• Differentiate the outer function with respect to its input, which is itself a function, using \( f'(f(f(f(x)))) \).
• Multiply by the derivative of each successive inner function, leading to the expression \( f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x) \) for \( g'(x) \).

The chain rule allows us to break down complex derivatives into manageable steps, by analyzing them layer by layer. It emphasizes how each function's rate of change contributes to the derivative of the composite function.

Differentiable Functions

Differentiable functions are those that possess a derivative at every point in their domain. This means they have a well-defined slope or rate of change, expressed mathematically. In our exercise, \( f \) is differentiable, meaning we can reliably find \( f'(x) \) at any \( x \).

Differentiability implies continuity, so a differentiable function won't have abrupt jumps or breaks. This characteristic is essential when applying the chain rule, as we need a smooth transition from one function application to the next. The derivative \( f'(x) \) truly represents the behavior of \( f \) locally around \( x \).

In the context of the exercise:

• The differentiable nature of \( f \) ensures that each composition of \( f \) inside \( g(x) \) remains differentiable.
• The calculation of \( g'(x) \) involves the product of several derivatives \( f'(f(x)) \), \( f'(f(f(x))) \), etc., each requiring \( f \) to be differentiable.

Differentiable functions form the backbone of many calculus operations and are crucial in ensuring the smooth, predictable manipulation of functions like those seen in our problem.

Function Composition

Function composition refers to the process of combining two or more functions where the output of one function becomes the input of another. This concept is at the heart of our exercise, where the function \( g(x) = f(f(f(f(x)))) \) is significantly composed of the iterated application of the function \( f \).

When dealing with function composition, it is important to:

• Understand how the functions interact with each other. Here, the repeated use of \( f \) builds a highly nested structure.
• Realize that the composite function \( g(x) \) will depend heavily on the properties of \( f \), such as its differentiability and derivative \( f'(x) \).
• Apply the chain rule to differentiate the composed functions successfully.

Each layer of function composition can add complexity. However, when carefully analyzed with techniques like the chain rule, they become manageable. Knowing how to handle compositions is vital for solving higher-level calculus problems.

Symmetry in Calculus

Symmetry in calculus often refers to similar patterns or properties that reappear under certain transformations, and these can greatly simplify our analysis or solutions. This exercise provides a perfect example of symmetry with the function \( f \) and its values at specific points.

Here's how symmetry appears in our problem:

• The conditions \( f(x_1) = x_2 \) and \( f(x_2) = x_1 \) create a sort of swap symmetry.
• The derivative equation \( g'(x_1) = (f'(x_1) \cdot f'(x_2))^2 \) and \( g'(x_2) = (f'(x_2) \cdot f'(x_1))^2 \) show symmetry through equal terms, differing only in their order.
• Since multiplication is commutative, \( a \cdot b = b \cdot a \), the symmetry in calculations results in \( g'(x_1) = g'(x_2) \), thus simplifying the process.

Recognizing symmetry can be a powerful tool in calculus, as it allows us to identify balanced relationships between parts of a function or equation. This recognition simplifies derivatives and often leads to elegant solutions, as seen in proving \( g'(x_1) = g'(x_2) \).