Average Rate Of Change: $4x^2 + 5/x^3$ On $[-1,3]$

When we talk about the average rate of change of a function over a specific interval, we're essentially asking for the average slope of the line that connects two points on the function's graph. Think of it as the overall trend or steepness of the function as you move from one point to another. It's a fundamental concept in calculus that helps us understand how quantities change over time or space. For our function, $g(x)=4 x^2+\frac{5}{x^3}$, and the interval $[-1,3]$, we're going to calculate this average rate of change. This involves finding the value of the function at the endpoints of the interval and then using a simple formula. It's like figuring out your average speed on a road trip: you divide the total distance traveled by the total time taken. In our case, the 'distance' is the change in the function's value, and the 'time' is the change in the input variable (x). So, let's dive into the specifics of how to calculate this for $g(x)=4 x^2+\frac{5}{x^3}$ on the interval $[-1,3]$ and see what that average slope turns out to be.

Understanding the Average Rate of Change Formula

The average rate of change of a function $f(x)$ over an interval $[a, b]$ is given by the formula:

$$\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}$$

This formula is derived directly from the slope formula for a line passing through two points $(a, f(a))$ and $(b, f(b))$. The numerator, $f(b) - f(a)$, represents the change in the function's output (the 'rise'), and the denominator, $b - a$, represents the change in the function's input (the 'run'). So, we're essentially calculating 'rise over run' for the secant line connecting the two points on the graph of the function. For our specific problem, the function is $g(x)=4 x^2+\frac{5}{x^3}$ and the interval is $[-1,3]$. This means our $a = -1$ and our $b = 3$. We need to find $g(-1)$ and $g(3)$ first, and then plug those values, along with $a$ and $b$, into the average rate of change formula. This process allows us to quantify how much the function's value changes, on average, for each unit increase in $x$ within the given interval. It's a crucial step before we even begin to think about instantaneous rates of change, which involve calculus's derivative.

Evaluating the Function at the Interval Endpoints

Before we can apply the average rate of change formula, we must first evaluate our function, $g(x)=4 x^2+\frac{5}{x^3}$, at the endpoints of our interval $[-1,3]$. Our interval's endpoints are $a = -1$ and $b = 3$.

Let's start with $a = -1$. We substitute $-1$ for every $x$ in the function:

$$g(-1) = 4(-1)^2 + \frac{5}{(-1)^3}$$

First, we calculate the powers: $(-1)^2 = 1$ and $(-1)^3 = -1$.

Now, substitute these values back into the equation:

$$g(-1) = 4(1) + \frac{5}{-1}$$

Performing the multiplication and division:

$$g(-1) = 4 - 5$$

$$g(-1) = -1$$

So, the value of the function at $x = -1$ is $-1$. This means one of the points on our graph is $(-1, -1)$.

Next, let's evaluate the function at $b = 3$. We substitute $3$ for every $x$ in the function:

$$g(3) = 4(3)^2 + \frac{5}{(3)^3}$$

Calculate the powers: $(3)^2 = 9$ and $(3)^3 = 27$.

Substitute these values back:

$$g(3) = 4(9) + \frac{5}{27}$$

Perform the multiplication:

$$g(3) = 36 + \frac{5}{27}$$

To add these, we need a common denominator, which is 27. We can rewrite 36 as $\frac{36 \times 27}{27} = \frac{972}{27}$.

$$g(3) = \frac{972}{27} + \frac{5}{27}$$

$$g(3) = \frac{977}{27}$$

So, the value of the function at $x = 3$ is $\frac{977}{27}$. This means the other point on our graph is $(3, \frac{977}{27})$. Now that we have the function's values at both endpoints, we are ready to calculate the average rate of change.

Calculating the Average Rate of Change

Now that we have evaluated our function at the interval's endpoints, $g(-1) = -1$ and $g(3) = \frac{977}{27}$, we can plug these values into the average rate of change formula:

$$\text{Average Rate of Change} = \frac{g(b) - g(a)}{b - a}$$

Here, $a = -1$, $b = 3$, $g(a) = g(-1) = -1$, and $g(b) = g(3) = \frac{977}{27}$.

Substitute these values:

$$\text{Average Rate of Change} = \frac{\frac{977}{27} - (-1)}{3 - (-1)}$$

Let's simplify the numerator and the denominator separately.

Numerator: $\frac{977}{27} - (-1) = \frac{977}{27} + 1$. To add 1, we express it as $\frac{27}{27}$:

$$\frac{977}{27} + \frac{27}{27} = \frac{977 + 27}{27} = \frac{1004}{27}$$

Denominator: $3 - (-1) = 3 + 1 = 4$.

Now, substitute these simplified values back into the formula:

$$\text{Average Rate of Change} = \frac{\frac{1004}{27}}{4}$$

To divide a fraction by a whole number, we can multiply the fraction by the reciprocal of the whole number (which is $\frac{1}{4}$):

$$\text{Average Rate of Change} = \frac{1004}{27} \times \frac{1}{4}$$

We can simplify this by dividing 1004 by 4. $1004 \div 4 = 251$.

$$\text{Average Rate of Change} = \frac{251}{27}$$

So, the average rate of change of the function $g(x)=4 x^2+\frac{5}{x^3}$ on the interval $[-1,3]$ is $\frac{251}{27}$. This value tells us that, on average, for every one unit increase in $x$ within the interval from -1 to 3, the function $g(x)$ increases by $\frac{251}{27}$ units. It's a positive average rate of change, indicating an overall upward trend of the function over this interval.

Interpretation of the Result

The average rate of change we calculated for $g(x)=4 x^2+\frac{5}{x^3}$ on the interval $[-1,3]$ is $\frac{251}{27}$. What does this number mean in practical terms? It's the slope of the secant line connecting the points $(-1, g(-1))$ and $(3, g(3))$ on the graph of $g(x)$. Since $\frac{251}{27}$ is a positive value (approximately 9.3), it indicates that, on average, the function $g(x)$ is increasing as $x$ increases from -1 to 3. If you were to draw a straight line between the point on the graph at $x=-1$ and the point on the graph at $x=3$, that line would have a slope of $\frac{251}{27}$. This doesn't mean the function is increasing at a constant rate throughout the entire interval – in fact, due to the $1/x^3$ term, the function has a vertical asymptote at $x=0$ and behaves quite differently on either side of it. However, the average rate of change gives us a single number that summarizes the overall trend. It's a useful metric for understanding the general behavior of a function over an interval, especially when comparing different functions or different intervals for the same function. A larger positive average rate of change would imply a steeper upward trend, while a negative value would suggest a downward trend.

Real-World Applications of Average Rate of Change

The concept of average rate of change is not just an abstract mathematical idea; it has numerous applications in the real world. For instance, consider calculating the average speed of a car. If a car travels 150 miles in 3 hours, its average speed is 150 miles / 3 hours = 50 miles per hour. This is the average rate of change of distance with respect to time. In finance, you might calculate the average annual return on an investment over several years. If an investment grew from $1000 to $1500 in 5 years, the average rate of change (average annual return) would be ($1500 - $1000) / 5 years = $100 per year, or a 10% average annual growth rate. In physics, it's used to describe average velocity or acceleration. In biology, it could represent the average growth rate of a population over a specific period. Even in everyday scenarios, like tracking your progress in a fitness program, you might look at your average weight loss per week over a month. The function $g(x)=4 x^2+\frac{5}{x^3}$ might model something like the cost of producing a certain number of items, where $x$ is the number of items. The average rate of change would then tell you the average change in cost per item produced over a production range. Understanding this concept helps us analyze and predict trends in various fields. For more insights into calculus and its applications, you can visit Khan Academy for excellent educational resources.