Problem 11 - Brief Applied Calculus

Question

Find the average rate of change of the given function between the following pairs of \(x\) -values. [Hint: See pages 95-96.] a. \(x=2\) and \(x=4\) b. \(x=2\) and \(x=3\) c. \(x=2\) and \(x=2.5\) d. \(x=2\) and \(x=2.1\) e. \(x=2\) and \(x=2.01\) f. What number do your answers seem to be approaching? $$ \underline{\phantom{xxx}} f(x)=2 x^{2}+x-2 $$

Step-by-Step Solution

Answer

The average rate of change approaches 16 as \( x_2 \) nears 2.

Step 1: Understanding the Problem

We need to find the average rate of change of the function \( f(x) = 2x^2 + x - 2 \) between two \( x \)-values. This is found using the formula \[ \frac{f(x_2) - f(x_1)}{x_2 - x_1} \].

Step 2: Calculate f(2) and f(4) for Part (a)

Evaluate the function at \( x = 2 \) and \( x = 4 \).\( f(2) = 2(2)^2 + 2 - 2 = 8 \), \( f(4) = 2(4)^2 + 4 - 2 = 34 \).Now find the average rate of change: \[ \frac{34 - 8}{4 - 2} = \frac{26}{2} = 13. \]

Step 3: Calculate f(2) and f(3) for Part (b)

Evaluate the function at \( x = 2 \) and \( x = 3 \).\( f(2) = 8 \) (previously calculated), \( f(3) = 2(3)^2 + 3 - 2 = 19 \).Find the average rate of change: \[ \frac{19 - 8}{3 - 2} = 11. \]

Step 4: Calculate f(2) and f(2.5) for Part (c)

Evaluate the function at \( x = 2.5 \).\( f(2.5) = 2(2.5)^2 + 2.5 - 2 = 15.25 \).Find the average rate of change: \[ \frac{15.25 - 8}{2.5 - 2} = 14.5. \]

Step 5: Calculate f(2) and f(2.1) for Part (d)

Evaluate the function at \( x = 2.1 \).\( f(2.1) = 2(2.1)^2 + 2.1 - 2 = 9.61 \).Find the average rate of change: \[ \frac{9.61 - 8}{2.1 - 2} = 16.1. \]

Step 6: Calculate f(2) and f(2.01) for Part (e)

Evaluate the function at \( x = 2.01 \).\( f(2.01) = 2(2.01)^2 + 2.01 - 2 \approx 8.0801 \).Find the average rate of change: \[ \frac{8.0801 - 8}{2.01 - 2} = 16.01. \]

Step 7: Analyzing the Trend for Part (f)

As \( x_2 \) approaches 2, the average rate of change approaches 16. Thus, our answers appear to be approaching 16.

Key Concepts

Calculus

Calculus is a powerful branch of mathematics that allows us to understand and describe change and motion. It is composed of two main parts: differential calculus and integral calculus.

• Differential calculus examines the process of change, focusing on derivatives and rates of change.
• Integral calculus revolves around the concept of accumulation, focusing on areas and volumes.

In this specific exercise, we utilized the average rate of change, which serves as an entry point into understanding derivatives. Calculus helps us break down complex problems into simpler parts, analyze them, and reassemble them into a comprehensive understanding. This foundational principle is what allows calculus to interpret and predict various phenomena in engineering and science effectively. By gaining fluency in calculus concepts, one can better tackle problems that involve continuous change, like those encountered in physics or economics.

Functions

Functions are the building blocks of calculus and mathematics in general. A function is a relationship between a set of inputs and a set of possible outputs, typically expressed as an equation. For each input, there is a unique output.
For instance, in the problem provided, we work with the function \(f(x) = 2x^2 + x - 2\). Here, \(x\) is the input variable, and \(f(x)\) is the output value or the result of the function for a given \(x\).

• The function shows how each input value maps to an output value based on the formula.
• Understanding how a function behaves is essential because it lets us predict outputs for any given input.

A central task in calculus is understanding how functions change over intervals, which informs us about their behavior. By analyzing functions, we can start to understand the changes in nature they model, such as motion, growth, or decay.

Derivatives

Derivatives are a core concept in calculus that describe the rate at which a function is changing at any given point. They represent how a function's output value changes as the input value changes.

• The process of finding a derivative is called differentiation.
• Geometrically, a derivative can be interpreted as the slope of the tangent line to the curve of the function at a given point.

In our exercise, by examining the average rate of change over smaller and smaller intervals, we were approaching the concept of a derivative. Mathematically, if we continue to narrow the interval until it becomes infinitesimally small, we find the instantaneous rate of change, which is precisely the derivative. Understanding derivatives enables mathematicians and scientists to model dynamic systems accurately, predict outcomes, and optimize processes across diverse fields such as physics, engineering, and economics.