Problem 34 - Calculus Early Transcendentals

Question

(a) Show that if \(f\) and \(g\) are functions for which $$ f^{\prime}(x)=g(x) \text { and } g^{\prime}(x)=f(x) $$ for all \(x,\) then \(f^{2}(x)-g^{2}(x)\) is a constant. (b) Show that the function \(f(x)=\frac{1}{2}\left(e^{x}+e^{-x}\right)\) and the function \(g(x)=\frac{1}{2}\left(e^{x}-e^{-x}\right)\) have this property.

Step-by-Step Solution

Answer

(a) \(f^2(x) - g^2(x)\) is constant. (b) \(f(x) = \frac{1}{2}(e^x + e^{-x})\) and \(g(x) = \frac{1}{2}(e^x - e^{-x})\) satisfy this.

Step 1: Differentiate the Expression

We are given two equations: \( f'(x) = g(x) \) and \( g'(x) = f(x) \). Let us consider the function \( h(x) = f^2(x) - g^2(x) \). To show \( h(x) \) is constant, we compute its derivative \( h'(x) \).Using the derivative rules, we have \( h'(x) = \frac{d}{dx} (f^2(x) - g^2(x)) = 2f(x)f'(x) - 2g(x)g'(x) \).

Step 2: Substitute Given Derivative Relations

Substitute the given relations into the derivative from Step 1:\( h'(x) = 2f(x)g(x) - 2g(x)f(x) \).Notice that the terms \( 2f(x)g(x) \) and \( -2g(x)f(x) \) cancel each other out, so \( h'(x) = 0 \).

Step 3: Conclude Constant Property

Since \( h'(x) = 0 \), it implies that \( h(x) \) is a constant function. Thus, \( f^2(x) - g^2(x) = C \) for some constant \( C \). This completes part (a), showing that \( f^2(x) - g^2(x) \) is indeed a constant.

Step 4: Verify the Given Functions

For part (b), we must show that the functions \( f(x) = \frac{1}{2}(e^x + e^{-x}) \) and \( g(x) = \frac{1}{2}(e^x - e^{-x}) \) satisfy the properties of part (a).First, compute \( f'(x): \frac{1}{2}(e^x - e^{-x}) = g(x) \), verifying \( f'(x) = g(x) \).

Step 5: Verify Second Given Derivative Relations

Next, compute \( g'(x): \frac{1}{2}(e^x + e^{-x}) = f(x) \), confirming \( g'(x) = f(x) \).Thus, both relations required for part (a) hold, and therefore \( f(x) \) and \( g(x) \) satisfy the properties.

Step 6: Conclusion: Verify Constant Property of Given Functions

Calculate \( f^2(x) - g^2(x) \) by substituting in the definitions of \( f(x) \) and \( g(x) \):\( f^2(x) = \frac{1}{4}(e^{2x} + 2 + e^{-2x}) \)\( g^2(x) = \frac{1}{4}(e^{2x} - 2 + e^{-2x}) \)Combining these gives \( f^2(x) - g^2(x) = 1 \), confirming the expression is indeed a constant.

Key Concepts

Derivatives

In calculus, a derivative represents the rate at which a function is changing at any given point. It is a core concept for understanding dynamic behaviors in mathematical functions and solving real-world problems involving motion, growth, and change. For example, if you are given a function like \( f(x) = x^2 \), its derivative \( f'(x) \) would be \( 2x \), indicating that the function's rate of change increases linearly as \( x \) increases.

Understanding derivatives is crucial in both analyzing and visualizing the behavior of functions.

• It helps us understand tangent lines, which are straight lines that just touch the curve at a specific point.
• They provide insights into the rates of change, helping to predict future behaviors of dynamics that the function represents.

In the problem provided, we are given two functions, \( f \) and \( g \), whose derivatives are linked in a unique way: \( f'(x) = g(x) \) and \( g'(x) = f(x) \). By computing the derivative, we unravel the relationship between these functions to determine other properties, such as constancy. When the derivative of a function (like \( h(x) = f^2(x) - g^2(x) \) in the problem) is zero, it means the function is constant over its domain.

Constant Function

A constant function is a fundamental concept in calculus where the output value remains the same no matter the input. This principle is evident when the derivative of a function is zero, indicating no change or variation in the function's output as the input changes.

In the given problem, our goal is to show that \( f^2(x) - g^2(x) \) is a constant function by illustrating that its derivative is zero.

• Firstly, compute \( h(x) = f^2(x) - g^2(x) \) and its derivative \( h'(x) \).
• After substituting the derivative relations \( f'(x) = g(x) \) and \( g'(x) = f(x) \), we find that \( h'(x) = 2f(x)g(x) - 2g(x)f(x) = 0 \).

Because the derivative \( h'(x) \) turns out to be zero, it clearly shows that the rate of change of \( h(x) \) is zero, meaning \( h(x) \) itself is constant. This verifies the premise that \( f^2(x) - g^2(x) = C \) for some constant \( C \). This property is fundamental to understanding when and why a function does not vary with changes in its input.

Exponential Functions

Exponential functions are of the form \( e^x \), where \( e \) is Euler's number, approximately 2.718. These functions are incredibly important in mathematics due to their property of growing very quickly. They are used extensively in various applications, from compound interest calculations to modeling population growth.

In the problem, the functions \( f(x) = \frac{1}{2}(e^x + e^{-x}) \) and \( g(x) = \frac{1}{2}(e^x - e^{-x}) \) exhibit a symmetrical and complementary nature.

• The derivative \( f'(x) = g(x) \) demonstrates their interconnectedness.
• Similarly, \( g'(x) = f(x) \) shows the reverse relationship.

These relationships indicate that exponential functions have derivatives that are intrinsically linked to the original function.

• This unique property is why exponential functions maintain their form upon differentiation.
• Aside from being used in theoretical contexts, they have practical applications in areas such as biology, demographics, and finance.

When these functions are squared and subtracted from each other, \( f^2(x) - g^2(x) \), it results in a constant value, demonstrating the rich structure and balance in exponential functions.