Problem 43 - Calculus with Concepts in Calculus

Question

a. Let \(f(x)=a x+b\). Show that \(f(x+1)-f(x)=a\). b. Let \(g(x)=b a^{x}\), where \(a\) is positive and \(b \neq 0 .\) Show that \(g(x+1) / g(x)=a\).

Step-by-Step Solution

Answer

f(x+1) - f(x) equals a; g(x+1)/g(x) equals a.

Step 1: Identify Functions

For both parts of the exercise, identify the functions given. In part (a), the function is linear: \(f(x) = ax + b\).In part (b), the function is exponential: \(g(x) = b a^x\).

Step 2: Calculate the Difference for f(x)

We need to show that \(f(x+1) - f(x) = a\) using the linear function from part (a).Calculate \(f(x+1)\):\[f(x+1) = a(x+1) + b = ax + a + b\]Calculate \(f(x+1) - f(x)\):\[f(x+1) - f(x) = (ax + a + b) - (ax + b) = a\]

Step 3: Verify the Result for f(x)

Review the differences calculated. The terms \(ax\) and \(b\) cancel out leaving the result as \(a\). Thus, \(f(x+1) - f(x) = a\) is indeed correct.

Step 4: Establish the Ratio for g(x)

We need to show that \(\frac{g(x+1)}{g(x)} = a\) using the exponential function from part (b).Calculate \(g(x+1)\):\[g(x+1) = b a^{x+1} = b a^x a\]Calculate the ratio \(\frac{g(x+1)}{g(x)}\):\[\frac{g(x+1)}{g(x)} = \frac{b a^x a}{b a^x} = a\]

Step 5: Verify the Result for g(x)

Check the ratio calculated. The terms \(b a^x\) cancel out, confirming the final result.Thus, \(\frac{g(x+1)}{g(x)} = a\) is also correct.

Key Concepts

Understanding Linear Functions

Linear functions are among the simplest types of functions you'll encounter in mathematics. They're characterized by their equation form, typically written as \( f(x) = ax + b \). This equation reveals two key features of a linear function: the slope, \( a \), and the y-intercept, \( b \).

• The slope \( a \) measures how steep the line is. If \( a \) is positive, the line slopes upward, while a negative \( a \) means it slopes downward.
• The y-intercept \( b \) is the point where the line crosses the y-axis.

Linear functions are called "linear" because when you graph them, they produce a straight line. This straight line reveals a constant rate of change – the same amount of change occurs for each step along the x-axis.

Exploring Exponential Functions

Exponential functions take a different route compared to linear functions. They are described by an equation of the form \( g(x) = ba^x \), where \( a \) is the base and \( b \) is a constant that affects the vertical stretch or compression. Here is what to know:

• The base \( a \) determines the growth or decay rate. If \( a \) is greater than 1, the function represents exponential growth. If \( a \) is between 0 and 1, it represents decay.
• The constant \( b \) modifies the initial value, effectively shifting the function up or down on the graph.

Unlike linear functions, the change in an exponential function increases or decreases at a rate proportional to its current value. This is why exponential functions have a characteristic curve, increasing slowly and then rapidly growing or decaying.

Determining Function Difference

When it comes to functions, particularly linear ones, the concept of function difference refers to the change in value when we substitute consecutive inputs. In other words, it's about finding how much one function value changes compared to the next.
For a linear function \( f(x) = ax + b \), you find the difference \( f(x+1) - f(x) \). By substituting, you arrive at:

• \( f(x+1) = a(x+1) + b \)
• \( f(x+1) - f(x) = ax + a + b - (ax + b) = a \)

This simplification shows that for linear functions, the amount added each step, \( a \), is constant. This constant difference is what makes the graph a straight line.

Calculating Function Ratio

The idea of a function ratio is notably important when discussing exponential functions. It involves comparing two function values at consecutive inputs, which helps understand growth or decay rates. For an exponential function \( g(x) = ba^x \), we are interested in calculating the ratio \( \frac{g(x+1)}{g(x)} \). Here’s the process:

• Start with \( g(x+1) = ba^{x+1} \).
• Then calculate the ratio: \( \frac{g(x+1)}{g(x)} = \frac{ba^{x+1}}{ba^x} \).
• This simplifies to \( a \), since \( ba^x \) in the numerator and denominator cancels out.

This result indicates that the function grows or decays by the factor of \( a \) each time \( x \) increases by 1, depending on the value of \( a \). This distinctive property makes studying exponential functions crucial in various fields such as finance and biology.