Problem 32 - Contemporary Precalculus

Question

Find the average rate of change of the function. \(f(x)=a^{x}, a>0,\) as \(x\) goes from 0 to 0.001

Step-by-Step Solution

Answer

Answer: \(\frac{a^{0.001} - 1}{0.001}\)

Step 1: Write the given function and the difference quotient formula

We have the function \(f(x) = a^x\) and the difference quotient formula: \(\frac{f(x_2) - f(x_1)}{x_2 - x_1}\)

Step 2: Substitute the given values into the formula

Using \(x_1 = 0\) and \(x_2 = 0.001\), plug these values into the difference quotient formula: \(\frac{f(0.001)-f(0)}{0.001-0}\)

Step 3: Write the function values for f(0) and f(0.001)

Find the function values for \(f(0)\) and \(f(0.001)\): \(f(0) = a^0 = 1\) and \(f(0.001) = a^{0.001}\)

Step 4: Plug the function values into the difference quotient formula

Substitute the function values found in step 3 into the difference quotient formula: \(\frac{a^{0.001} - 1}{0.001}\)

Step 5: Simplify the expression

We can't simplify the expression further unless we are given a specific value for \(a\). Therefore, the average rate of change of the function as \(x\) goes from 0 to 0.001 is: \(\frac{a^{0.001} - 1}{0.001}\)

Key Concepts

Difference Quotient

The difference quotient is a fundamental tool in calculus. It measures how a function changes over a small interval. Think of it as the average rate of change or slope of a function between two points. The formula is:

• \(\frac{f(x_2) - f(x_1)}{x_2 - x_1}\)

This formula calculates the difference in the function values \(f(x_2)\) and \(f(x_1)\) (which can be thought of as the "rise"), divided by the difference in the input values \(x_2\) and \(x_1)\) (the "run").
For example, when we consider our specific problem:

• \(x_1 = 0\) and \(x_2 = 0.001\)

We apply these to the formula to find the rate of change over that tiny interval. This step is crucial to grasping more complex ideas in calculus, such as derivatives.

Exponential Functions

Exponential functions are functions where a constant base \(a\) is raised to a variable exponent \(x\). They are written in the form \(f(x) = a^x\), where \(a\) is a positive constant.
They are characterized by rapid growth or decay, depending on whether \(a > 1\) (growth) or \(0 < a < 1\) (decay).
This type of function is common in many real-world scenarios, such as population growth, finance compound interest, and radioactive decay.

• In the exercise, the function given is \(f(x) = a^x\).
• At \(x=0\), the exponential function simplifies to \(f(0) = a^0 = 1\). This is because any non-zero number raised to the power of 0 is 1.
• Increasing the exponent slightly, as with \(x=0.001\), gives us \(f(0.001) = a^{0.001}\).

Understanding these functions helps in solving the problem by determining the function values needed for the difference quotient.

Function Evaluation

Function evaluation involves plugging in specific input values into a function to get corresponding outputs. This is a vital step in using the difference quotient to estimate the average rate of change.
For the exponential function \(f(x) = a^x\):

• The given task is to determine the function values for specific \(x\) values: \(x = 0\) and \(x = 0.001\).
• At \(x = 0\), \(f(0) = a^0 = 1\).
• At \(x = 0.001\), \(f(0.001) = a^{0.001}\). This slight change in \(x\) results in a very slight change in the function's output due to the smallness of the exponent.

Function evaluation is straightforward but pivotal in applying more complex calculus concepts. It makes sure we accurately compute function-related changes and apply them to mathematical models.