Problem 84 - Calculus Early Transcendentals: Pearson New International Edition

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 \(g^{\prime}\left(x_{1}\right)=g^{\prime}\left(x_{2}\right)\).

Step-by-Step Solution

Answer

Given symmetrical properties of repeated function compositions, \( g'(x_1) = g'(x_2) \).

Step 1: Differentiate g(x)

Start by differentiating the given function \( g(x) = f(f(f(f(x)))) \) using the chain rule. Applying the chain rule, we have: \[ g'(x) = f'(f(f(f(x)))) \cdot f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x) \] This expression represents the derivative of \( g(x) \), which comprises several compositions and derivatives of \( f(x) \).

Step 2: Evaluate g'(x_1)

Substitute \( x = x_1 \) in the derivative expression obtained: \[ g'(x_1) = f'(f(f(f(x_1)))) \cdot f'(f(f(x_1))) \cdot f'(f(x_1)) \cdot f'(x_1) \] Since \( f(x_1) = x_2 \), substitute \( x_2 \) into the equation: \[ g'(x_1) = f'(f(f(x_2))) \cdot f'(f(x_2)) \cdot f'(x_2) \cdot f'(x_1) \]

Step 3: Evaluate g'(x_2)

Substitute \( x = x_2 \) in the derivative expression: \[ g'(x_2) = f'(f(f(f(x_2)))) \cdot f'(f(f(x_2))) \cdot f'(f(x_2)) \cdot f'(x_2) \] Since \( f(x_2) = x_1 \), substitute \( x_1 \) into the equation: \[ g'(x_2) = f'(f(f(x_1))) \cdot f'(f(x_1)) \cdot f'(x_1) \cdot f'(x_2) \]

Step 4: Compare g'(x_1) and g'(x_2)

Now, compare \( g'(x_1) \) and \( g'(x_2) \): Both expressions involve the same components rearranged, as one starts with \( f(f(f(x_2))) \) and the other with \( f(f(f(x_1))) \), and each progresses through the composition in the reverse order. Note that for functions \( f \) and its compositions, each substitution results symmetrically in reversed sequences under given conditions. Since both expressions are identical apart from their rearrangement, we have: \[ g'(x_1) = g'(x_2) \]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used for differentiating composite functions. At its core, it enables us to find the derivative of a function that is built up from other functions. Think of it as a recipe where each step depends on the result of the previous one. So, if you have a function like

• \( h(x) = f(g(x)) \)

then, to find its derivative \( h'(x) \), you multiply the derivative of the outer function by the derivative of the inner function:\( h'(x) = f'(g(x)) \, imes \, g'(x) \)This is what we applied in step 1 of the solution, where the function \( g(x) = f(f(f(f(x)))) \) was differentiated. Each layer of \( f \) contributes its own derivative to the final expression, resulting in a product of derivatives.

Composite Functions

Composite functions are essentially a combination of functions where the output of one function becomes the input of another. For example, in this problem, the given function is

• \( g(x) = f(f(f(f(x)))) \)

This function is a composition of the function \( f \) applied four times. Imagine it as a layering process, where each application of \( f \) forms a new layer over the previous function.Composite functions can be tricky because they involve multiple interconnected pieces. When handling such functions, you'll often use the chain rule to differentiate them. This process involves meticulously treating each part of the function and finding its derivative.

Differentiation

Differentiation is the mathematical process of finding the rate at which a function changes at any point. It's like figuring out how fast your car is going at any moment in time. For any differentiable function \( f \), the derivative \( f'(x) \) gives you the slope of the function at a given point. In our original problem, the differentiation is achieved by using the chain rule to handle the complex composite function. We start with the outermost function and work our way inwards.

• Keep in mind, the application of differentiation here not only allows us to obtain the derivative but also ensures that our numerical manipulations align with the rules of calculus.

Symmetry in Functions

Symmetry in functions often implies that the function exhibits a form of balance or regularity. In the context of this problem, we need to show that

• \( g'(x_1) = g'(x_2) \)

where both points are interconnected through the function \( f \).Symmetry here doesn't mean visual equality but functional equivalence under certain conditions. Each of the derivatives, \( g'(x_1) \) and \( g'(x_2) \), contains similar components, but in reverse order. This reversal stems directly from the relationships:

• \( f(x_1) = x_2 \)
• \( f(x_2) = x_1 \)

Thus, when we substitute and simplify, we find these expressions produce the same result, demonstrating symmetry.