Problem 3 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(1-6,\) find the average rate of change of the function over each interval. \(f(x)=e^{x}\) (a) $$[-2,0] \quad$$ (b) $$[1,3]$$

Step-by-Step Solution

Answer

The average rate of change over the interval \([-2,0]\) is \(\frac{1 - e^{-2}}{2}\) and over the interval \([1,3]\) is \(\frac{e^{3} - e^{1}}{2}\)

Step 1: Average rate of change on interval [-2,0]

Apply the formula for the average rate of change to the first interval \([-2,0]\) in the function \(f(x)=e^{x}\). Here, \(f(-2) = e^{-2}\) and \(f(0) = e^{0}\). Therefore, the average rate of change for this interval is \(\frac{f(0) - f(-2)}{0-(-2)} = \frac{e^{0} - e^{-2}}{0-(-2)}\)

Step 2: Simplify

Simplify the expression from step 1 to get a number. Since \(e^{0}\) equals 1 and \(0-(-2)\) equals 2, the average rate of change for this interval is \(\frac{1 - e^{-2}}{2}\)

Step 3: Average rate of change on interval [1,3]

Apply the formula for the average rate of change to the second interval \([1,3]\) in the function \(f(x)=e^{x}\). Here, \(f(1) = e^{1}\) and \(f(3) = e^{3}\). Therefore, the average rate of change for this interval is \(\frac{f(3) - f(1)}{3-1} = \frac{e^{3} - e^{1}}{3-1}\)

Step 4: Simplify

Simplify the expression from step 3 to get a number. Since \(3-1\) equals 2, the average rate of change for this interval is \(\frac{e^{3} - e^{1}}{2}\)

Key Concepts

Exponential function

An exponential function is a type of mathematical function where the variable appears in the exponent. The general form is \( f(x) = a \cdot b^{x} \), where \( a \) and \( b \) are constants, and \( b \) is a positive real number. In our case, the function is \( f(x) = e^{x} \), making it a natural exponential function.

• \(e \approx 2.718\) is the base of the natural logarithm, usually used in continuous growth models.
• Exponential functions are defined for all real numbers and have a constant growth rate called the "rate of change".

Understanding the exponential function is vital for solving problems involving rates of growth, like interest calculations and population models.

Intervals

Intervals on the real line can be thought of as a continuous range of numbers. In function analysis, they are used to define specific start and endpoints. For this exercise:- An interval is denoted using brackets such as \([-2,0]\) or \([1,3]\).- The square brackets \([a,b]\) mean the interval includes both \(a\) and \(b\), referred to as a closed interval.To calculate the average rate of change of a function over an interval, you evaluate the function at the endpoints of the interval and apply these evaluations in the rate of change formula. This helps us understand how the function behaves over a specific range.

Function evaluation

Function evaluation is the process of determining the output of a function using specific input values. For example, to evaluate the function \(f(x) = e^{x}\) at \(x = -2\), substitute \(x\) with \(-2\) to get \(f(-2) = e^{-2}\). Similarly, evaluate at \(x = 0\) to get \(f(0) = e^{0}\).

• This process is essential for solving problems involving rates of change and areas under curves, as it provides necessary values for further calculations.
• The results from evaluating functions are used in subsequent steps like simplification or determining the rate of change.

In exponential functions, computation often involves transforming the base \(e\) raised to different powers.

Simplification of expressions

Simplification involves reducing complicated expressions into simpler forms for ease of interpretation and calculation. By substituting known values, like \( e^{0} = 1 \), we can break down expressions. For instance, the expression \(\frac{e^{0} - e^{-2}}{2}\) simplifies because:

• \(e^{0}\) is simplified to 1, as any number raised to the power of 0 equals 1.
• The subtraction \(1 - e^{-2}\) comes from exponent rules and needs calculation or estimation for further simplification.

Simplifying these expressions makes complex function evaluations more manageable and prepares them for interpretation or solving further problems.