Problem 46 - Precalculus with Calculus Previews

Question

For \(f(x)=b^{x},\) show that: (a) \(\frac{f(x+h)-f(x)}{h}=b^{x}\left(\frac{b^{h}-1}{h}\right)\) (b) \(f\left(x_{1}+x_{2}\right)=f\left(x_{1}\right) f\left(x_{2}\right)\)

Step-by-Step Solution

Answer

(a) It simplifies to \( b^{x} \cdot \frac{b^{h}-1}{h} \). (b) \( f(x_1 + x_2) = f(x_1)f(x_2) \).

Step 1: Define the Function Components for Part (a)

The function given is \( f(x) = b^x \). To find \( \frac{f(x+h)-f(x)}{h} \), we express the change as \( f(x+h) = b^{x+h} \). Thus, we compare it with \( f(x) = b^x \).

Step 2: Simplify the Difference Quotient for Part (a)

The difference quotient is \( \frac{f(x+h) - f(x)}{h} = \frac{b^{x+h} - b^x}{h} \). This can be rewritten using the property \( b^{x+h} = b^x \cdot b^h \). So, the expression becomes \( \frac{b^x \cdot b^h - b^x}{h} = \frac{b^x (b^h - 1)}{h} \).

Step 3: Finalize the Solution for Part (a)

Factor \( b^x \) out of the fraction, resulting in \( b^x \cdot \frac{b^h - 1}{h} \). This agrees with the expression given in part (a).

Step 4: Define the Function Components for Part (b)

For \( f(x_1 + x_2) \), we begin by recognizing that \( f(x) = b^x \). Hence, \( f(x_1 + x_2) = b^{x_1 + x_2} \).

Step 5: Simplify Using Exponential Properties for Part (b)

Use the exponential property \( b^{x_1 + x_2} = b^{x_1} b^{x_2} \). Thus, \( f(x_1 + x_2) = b^{x_1} b^{x_2} \), which is equivalent to \( f(x_1)f(x_2) \). This completes the derivation for part (b).

Key Concepts

Difference Quotient

The difference quotient is a central concept in calculus as it forms the foundation for understanding derivatives. It measures the average rate of change of a function over a small interval. For the function given as \(f(x) = b^x\), the difference quotient calculated for the increment \(h\) is expressed as \(\frac{f(x+h) - f(x)}{h}\). This is particularly challenging for exponential functions due to their unique properties.

To solve this, we need to express \(f(x+h)\) in terms of known quantities. For the exponential function \(f(x)=b^x\), the value at \(x+h\) becomes \(b^{x+h}\). With the help of the exponential property \(b^{x+h} = b^x \cdot b^h\), the difference quotient can be transformed into:

\[\frac{b^x \cdot b^h - b^x}{h} = \frac{b^x (b^h - 1)}{h}\]

Here, by factoring \(b^x\) out of the numerator, we simplify the expression to \(b^x \left( \frac{b^h - 1}{h} \right)\). This represents the rate of change of the exponential function and highlights the interplay between the exponential base and the rate of change in its vicinity.

Exponential Properties

Exponential properties provide key insights into how functions involving exponents operate. Exponentiation is a powerful mathematical tool that follows distinct laws like product of powers, quotient of powers, and power of a power. These properties simplify complex algebra involving exponential functions.

In our exercise, one key property used is that \(b^{x_1 + x_2} = b^{x_1} b^{x_2}\). This property states that multiplying exponential terms with the same base results in adding their exponents. Hence, whether you're dealing with exponential growth or decay, these properties maintain coherence across various manipulations.

Specifically, in part (b) of our exercise, you can see how this property lets us break down \(f(x_1 + x_2)\) as \(f(x_1) f(x_2)\). Writing it as \(b^{x_1 + x_2} = b^{x_1} b^{x_2}\), integrally demonstrates the utility of these exponential laws in simplifying and solving expressions efficiently.

Function Composition

Function composition involves applying one function to the results of another. It's a way to create complex functions from simpler ones. In terms of exponential functions, the concept of function composition results in manipulating input attributes to achieve desired outputs.

In our context, examining the function \(f(x_1 + x_2)\) can be viewed as a type of function composition. Although not a direct composition of functions in usual terms, structuring it to match \(f(x_1)f(x_2)\) exploits properties of exponents that mirror what happens when functions are composed.

This highlights an important aspect of functional analysis, where combining function rules or properties can often achieve results comparable to certain compositions. Understanding these scenarios builds a strong foundation for more advanced mathematical explorations.