Problem 52 - Calculus Early Transcendentals

Question

Suppose that a function \(f\) is differentiable at \(x_{0}\) and define \(g(x)=f(m x+b),\) where \(m\) and \(b\) are constants. Prove that if \(x_{1}\) is a point at which \(m x_{1}+b=x_{0},\) then \(g(x)\) is differentiable at \(x_{1}\) and \(g^{\prime}\left(x_{1}\right)=m f^{\prime}\left(x_{0}\right)\)

Step-by-Step Solution

Answer

Yes, \( g(x) \) is differentiable at \( x_1 \) and \( g'(x_1) = m f'(x_0) \).

Step 1: Understanding the Problem

We are given a function \( f \) differentiable at a point \( x_0 \). The goal is to prove that the function \( g(x) = f(mx + b) \) is differentiable at a point \( x_1 \) such that \( mx_1 + b = x_0 \), and its derivative \( g'(x_1) = m f'(x_0) \).

Step 2: Applying the Chain Rule

The function \( g(x) = f(mx + b) \) can be considered as a composition of two functions: the inner function \( h(x) = mx + b \) and the outer function \( f(u) \), where \( u = h(x) = mx + b \). According to the chain rule, the derivative \( g'(x) \) is given by \( g'(x) = f'(u) \cdot h'(x) \).

Step 3: Calculating the Derivative of the Inner Function

The derivative of the inner function \( h(x) = mx + b \) is \( h'(x) = m \), since the derivative of a linear term is the coefficient of \( x \). The constant \( b \) has a derivative of 0.

Step 4: Evaluating at Specific Points

We know that at \( x_1 \), \( mx_1 + b = x_0 \). Therefore, when we compute the derivative \( g'(x_1) = f'(x_0) \cdot m = m f'(x_0) \), since \( u = h(x_1) = mx_1 + b = x_0 \).

Step 5: Conclusion of the Proof

Since \( g(x) \) is differentiable at \( x_1 \) via the chain rule evaluation and yields the result \( g'(x_1) = m f'(x_0) \), we have proved the required statement using calculus concepts.

Key Concepts

Chain Rule

The Chain Rule is a fundamental technique in calculus for finding the derivative of a composite function. When you have a function that is the result of one function applied to another, the chain rule helps you differentiate it.

Consider the function you are given in the original exercise:

\( g(x) = f(mx + b) \).

This function can be viewed as a composition of two functions. To find the derivative \(g'(x)\), the chain rule is applied. First, differentiate the outer function while keeping the inner function intact. Then, multiply the result by the derivative of the inner function. The chain rule is a handy tool because:

• It reduces complex differentiation problems into simpler parts.
• It's crucial for functions with nested forms, like \( g(x) = f(mx + b) \).

By breaking down the function into manageable pieces, the chain rule streamlines the process of finding derivatives, aiding in understanding complex relationships between functions.

Composition of Functions

The composition of functions is an essential concept in understanding how functions interact with each other to create new functions.

In the given problem, we look at \( g(x) = f(mx + b) \), where:

• The inner function is \( h(x) = mx + b \).
• The outer function is \( f(u) \), where \( u = h(x) \).

Through composition, we plug the result of one function into another.

This process allows for building more complex functions from simpler ones while retaining an understanding of each piece. In differentiable terms, composition ensures that the behavior and rate of change of the function can be understood by examining each part's derivative.

The composition of functions is like connecting pipes in a plumbing system; each section can be examined for performance individually, yet they all contribute to the system's overall functionality.

Calculus Proof

The exercise solution involves employing a calculus proof strategy to demonstrate function differentiability. Calculus proofs often require validating how one statement logically follows from others, using fundamental calculus principles.

Here, the aim is to show that \( g(x) = f(mx + b) \) is differentiable at some point \( x_1 \) and to find its derivative \( g'(x_1) \).

• Using previously covered theorems such as the chain rule helps streamline the proof process.
• Explicitly calculate derivatives to ensure every step is logically sound.

This proof shows that the derivative \( g'(x_1) = m \cdot f'(x_0) \) is not just a coincidental result, but a necessary consequence of the rules of differentiation.

The meticulousness in each step ensures that the conclusion is valid. In essence, the calculus proof confirms hypotheses about function behavior, providing clear evidence that emerges from precise mathematical reasoning.