Problem 52 - Calculus - AP Edition

Question

Suppose that a function \(f\) is differentiable at \(x_{0}\) and \(\mathrm{de}\) fine \(g(x)=f(m x+b),\) 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 \( x_1 \) with \( g'(x_1) = m f'(x_0) \).

Step 1: Understand the Definitions

We are given a function \( f \) that is differentiable at \( x_0 \). The function \( g(x) = f(mx + b) \). We need to prove that \( g(x) \) is differentiable at \( x_1 \) where \( mx_1 + b = x_0 \), and show that \( g'(x_1) = m f'(x_0) \).

Step 2: Recognize the Condition for Differentiability

Since \( f \) is differentiable at \( x_0 \), by definition, the derivative is \( f'(x_0) = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h} \). Our task is to express a similar limit definition for the differentiability of \( g \) at \( x_1 \).

Step 3: Set Up the Expression For \( g'(x) \)

Using the chain rule of differentiation, \( g(x) = f(mx + b) \) leads to: \[ g'(x) = \frac{d}{dx}[f(mx + b)] = f'(mx + b) \cdot \frac{d}{dx}[mx + b]. \] Since \( \frac{d}{dx}[mx + b] = m \), we have \( g'(x) = m f'(mx + b) \).

Step 4: Evaluate Derivative at \( x_1 \)

Plug in \( x_1 \) into the expression for \( g'(x) \): \[ g'(x_1) = m f'(mx_1 + b). \] Given \( mx_1 + b = x_0 \), we substitute to find: \[ g'(x_1) = m f'(x_0). \]

Step 5: Conclude Differentiability and Derivative Result

Since we have expressed \( g'(x_1) \) explicitly in terms of \( f'(x_0) \), this proves that \( g(x) \) is differentiable at \( x_1 \), and importantly, that \( g'(x_1) = m f'(x_0) \), consistent with the chain rule.

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus that helps us differentiate composite functions. A composite function is essentially a function within a function. Think of it as a "nested" function structure, like Russian dolls where one fits inside another. When dealing with these types of functions, direct differentiation isn't straightforward, so we use the Chain Rule.

To apply the Chain Rule, consider a function of the form \( g(x) = f(u(x)) \), where \( u(x) \) is an inner function and \( f(u) \) is the outer function. The Chain Rule states that the derivative of \( g(x) \) with respect to \( x \) is the derivative of \( f \) with respect to \( u \), multiplied by the derivative of \( u \) with respect to \( x \). In notation, this is expressed as:

• \( g'(x) = f'(u) \cdot u'(x) \).

This method was employed in the original exercise to differentiate \( g(x) = f(mx + b) \), where \( mx + b \) acts as the inner function, allowing us to find the derivative of \( g(x) \) given the derivative of \( f(x) \) and the constants \( m \) and \( b \).

Function Composition

Function Composition combines two functions to create a new function. When we compose two functions, say \( f(x) \) and \( g(x) \), the result is a function \( h(x) = f(g(x)) \). Here, \( g(x) \) becomes the input for \( f(x) \).

Function composition can sometimes make the process of differentiation tricky. That's where tools like the Chain Rule come into play. Composition often appears in practical applications where one process depends on another. In our current exercise, \( g(x) = f(mx + b) \) is a composition: the linear function \( mx + b \) is nested within \( f(x) \). Understanding this relationship helps us effectively use differentiation rules.

Breaking down a composite function into its simpler components can make differentiation clearer. This approach highlights the importance of identifying the inner and outer functions, which is essential for applying methods like the Chain Rule to differentiate effectively.

Derivative Calculation

Calculating derivatives is a core aspect of calculus, focusing on finding the rate at which a function changes. The derivative provides crucial information about the function's behavior, such as its slope at a point or the rate of change.

The process typically involves applying differentiation rules systematically. For linear functions, the derivative is the constant coefficient, while for more complex functions, a combination of rules might be needed. In our exercise, the given function \( g(x) = f(mx + b) \) uses the known derivative of the inner function \( mx + b \), which is \( m \), alongside the derivative of the outer function \( f \) with respect to its input.

To find \( g'(x) \), the Chain Rule guided us to multiply \( f'(mx + b) \) by the derivative of \( mx + b \), yielding \( m f'(mx + b) \). Calculating this at the specific point \( x_1 \), where \( mx_1 + b = x_0 \), confirmed that \( g'(x_1) = m f'(x_0) \). Such calculations show how derivatives can convey essential details about function changes, useful in various disciplines like physics and engineering.