Problem 10 - 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. \(x=1\) and \(x=3\) b. \(x=1\) and \(x=2\) c. \(x=1\) and \(x=1.5\) d. \(x=1\) and \(x=1.1\) e. \(x=1\) and \(x=1.01\) f. What number do your answers seem to be approaching? $$ \underline{\phantom{xxx}} f(x)=2 x^{2}+5 $$

Step-by-Step Solution

Answer

The average rate of change seems to approach 4 as the x-values get closer.

Step 1: Understanding the Problem

We are tasked with finding the average rate of change of the function \(f(x) = 2x^2 + 5\) over different intervals. The average rate of change of a function between two points \(x=a\) and \(x=b\) can be found using the formula: \[\frac{f(b) - f(a)}{b - a}\].

Step 2: Finding the Average Rate of Change for a

For \(x=1\) and \(x=3\), calculate the function values: \(f(1) = 2(1)^2 + 5 = 7\) and \(f(3) = 2(3)^2 + 5 = 23\). The average rate of change is \[\frac{f(3) - f(1)}{3 - 1} = \frac{23 - 7}{2} = 8\].

Step 3: Finding the Average Rate of Change for b

For \(x=1\) and \(x=2\), calculate the function values: \(f(2) = 2(2)^2 + 5 = 13\). The average rate of change is \[\frac{f(2) - f(1)}{2 - 1} = \frac{13 - 7}{1} = 6\].

Step 4: Finding the Average Rate of Change for c

For \(x=1\) and \(x=1.5\), calculate the function values: \(f(1.5) = 2(1.5)^2 + 5 = 9.5\). The average rate of change is \[\frac{f(1.5) - f(1)}{1.5 - 1} = \frac{9.5 - 7}{0.5} = 5\].

Step 5: Finding the Average Rate of Change for d

For \(x=1\) and \(x=1.1\), calculate the function values: \(f(1.1) = 2(1.1)^2 + 5 \approx 7.42\). The average rate of change is \[\frac{f(1.1) - f(1)}{1.1 - 1} \approx \frac{7.42 - 7}{0.1} \approx 4.2\].

Step 6: Finding the Average Rate of Change for e

For \(x=1\) and \(x=1.01\), calculate the function values: \(f(1.01) = 2(1.01)^2 + 5 \approx 7.0402\). The average rate of change is \[\frac{f(1.01) - f(1)}{1.01 - 1} \approx \frac{7.0402 - 7}{0.01} \approx 4.02\].

Step 7: Observation as x-values approach each other

As the intervals between x-values decrease from the initial x-value, the average rate of change is approaching the derivative at \(x=1\). Calculate it using the derivative \(f'(x) = 4x\) which gives \(f'(1) = 4\).

Key Concepts

Quadratic Functions

Quadratic functions are a special kind of polynomial function. They are often written in the form of \( f(x) = ax^2 + bx + c \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \). These functions create a U-shaped curve known as a parabola when graphed.
Parabolas can open upwards or downwards depending on the sign of the coefficient \( a \). If \( a > 0 \), the parabola opens upwards, and if \( a < 0 \), it opens downwards. The highest or lowest point of the parabola is called the vertex.
Understanding the shape and properties of quadratic functions is crucial in calculus since they form the basis for many real-world scenarios, like projectile motion or maximizing areas. Knowing how to find average rates of change with quadratic functions helps in understanding how quickly something is changing over time.

Derivatives

Derivatives are a core concept in calculus, representing how a function changes as its input changes.
The derivative of a function at a point measures the "instantaneous rate of change" or "slope" of the function at that exact spot. To keep it simple, it tells us how steep the curve is at any particular point.

• For the function \( f(x) = 2x^2 + 5 \), the derivative \( f'(x) \) is found using basic calculus rules to be \(4x\).
• The derivative \( f'(x) = 4x \) indicates that the slope of the function depends directly on \( x \).
• As a result, the slope increases as \( x \) increases, and it decreases as \( x \) decreases.

This is particularly relevant when finding the average rate of change across intervals and noticing that as you shrink the interval smaller and smaller, you approach the derivative.

Calculus

Calculus is a branch of mathematics focused on change. It deals with the study of how things change and accumulate over time or space. This is primarily done using two main tools: derivatives and integrals.
When we talk about the average rate of change in the context of calculus, we're looking at how a function behaves over a specified interval. This is like observing the average speed of a car over a journey.

• The average rate of change of the function \( f(x) = 2x^2 + 5 \) over different intervals gives us insight into how the function behaves as a whole.
• By calculating this rate over smaller and smaller intervals around a point, we can see how the function is changing precisely at that point.
• This leads to understanding derivatives, as they are the limit of these average rate calculations as the interval approaches zero.

Thus, calculus provides a deeper understanding of dynamic systems in the natural world, allowing us to predict and explain real-life phenomena.