Problem 66 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose that \(f(x+y)=f(x) f(y)\) for all \(x\) and \(y\). Show that if \(f^{\prime}(0)\) exists then \(f^{\prime}(a)\) exists and \(f^{\prime}(a)=f(a) f^{\prime}(0)\).

Step-by-Step Solution

Answer

If \( f'(0) \) exists, then \( f'(a) = f(a)f'(0) \).

Step 1: Understand the Given Functional Equation

The given equation is a functional equation: \( f(x+y) = f(x)f(y) \). This suggests that the function \( f \) could be of the form that respects properties of exponentials since the addition of inputs results in multiplication of outputs.

Step 2: Evaluate the Function at Zero

Consider the value of the function at zero by substituting \( x = 0 \) and \( y = 0 \) into the equation: \( f(0+0) = f(0)f(0) \), which implies \( f(0)^2 = f(0) \). Thus, \( f(0) = 1 \) if \( f(0) eq 0 \).

Step 3: Differentiate the Functional Equation

Differentiate both sides of \( f(x+y) = f(x)f(y) \) with respect to \( x \). Using the chain rule, we get: \( \frac{d}{dx}f(x+y) = f'(x+y), \) and the product rule gives \( f'(x)f(y) \). Thus, \( f'(x+y) = f'(x)f(y) \).

Step 4: Evaluate at y = 0

Set \( y = 0 \) in the differentiated equation: \( f'(x+0) = f'(x)f(0) \). Knowing \( f(0) = 1 \), it simplifies to \( f'(x) = f'(x) \), a tautology, ensuring consistency.

Step 5: Establish the Relationship for f'(a)

To show \( f'(a) = f(a)f'(0) \), consider the definition of the derivative: \( f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \). Using the original functional equation, \( f(a+h) = f(a)f(h) \).

Step 6: Calculate the Limit

Substitute into the derivative expression: \( f'(a) = \lim_{h \to 0} \frac{f(a)f(h) - f(a)}{h} = \lim_{h \to 0} f(a) \frac{f(h) - 1}{h} \). Factoring out \( f(a) \), we have \( f'(a) = f(a) \lim_{h \to 0} \frac{f(h) - 1}{h} = f(a)f'(0) \), since \( f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h) - 1}{h} \).

Key Concepts

Calculus Derivatives

Calculus derivatives are a fundamental concept in mathematics used to understand how a function changes as its input changes. In simple terms, a derivative tells us the rate at which a function's value is changing at any given point. Imagine riding a bicycle down a hill. The speed at which you are traveling at any moment is like the derivative of your journey.

When we find a derivative, we are essentially trying to find the slope of the tangent line to the curve of the function at a specific point. For a linear function, the slope is constant, while for more complex functions, it can vary at every point. This is where the derivative comes in handy.

• Tangent Lines: These are straight lines that touch a curve at a single point without crossing it. The slope of this line at a specific point is what we find by differentiating the function.
• Instantaneous Rate of Change: Derivatives help calculate how one quantity changes in relation to another instantly.

Derivatives involve rules like product rule, chain rule, and quotient rule, which aid in finding derivatives of various types of functions, including the one given in our problem, establishing important relationships and solving challenging calculus problems.

Exponential Functions

Exponential functions are types of functions where the variable is in the exponent, making them grow very quickly. This kind of function is written in the form \(f(x) = a^x\), where \(a\) is a positive constant. In our functional equation \(f(x+y) = f(x)f(y)\), we observe a characteristic reminiscent of exponential functions.

These functions are crucial in many real-world scenarios because they model growth and decay, such as population growth, radioactive decay, and compound interest.

• Growth and Decay: Real-world phenomena like population growth or chemical reactions often follow exponential patterns.
• Base of the Exponential: The base \(a\) determines how fast the function grows or decays.

The formula \(f(x+y) = f(x)f(y)\) hints that if \(f(0) = 1\), then \(f(x)\) could be an exponential function. Such functions help achieve solutions to complex equations by simplifying the way we manipulate and understand growth trends.

Limit Definition of Derivative

The limit definition of a derivative is a foundational concept in calculus. It formalizes how to find the derivative of a function at any given point. This definition is expressed as:

\[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]
It helps to understand the behavior of functions as we approach a particular input, capturing the slope or instantaneous rate of change at a point.

• Approaching Zero: The idea behind this definition is to observe how the difference quotient \( \frac{f(a+h) - f(a)}{h} \) behaves as \(h\) becomes infinitely small (approaches zero).
• Smooth Transition: The limit ensures that we measure the function's behavior very precisely near the point \(a\).

In the context of the exercise given, the limit definition was pivotal. It allows us to relate the derivative at a given point \( a \) with the derivative at zero which gives us the solution: \( f'(a) = f(a)\,f'(0) \). This relational power provides deep insights into how functions alter across their domains.