Problem 13 - Contemporary Precalculus

Question

Find the average rate of change of the function f over the given interval. $$f(x)=x^{3}-3 x^{2}-8 x+6 \text { from } x=-1 \text { to } x=3$$

Step-by-Step Solution

Answer

Answer: -7

Step 1: Find the function values at x = -1 and x = 3

To find the function values at x = -1 and x = 3, we will substitute the respective x-values into the given function: f(x) = x^3 - 3x^2 - 8x + 6 f(-1) = (-1)^3 - 3(-1)^2 - 8(-1) + 6 f(3) = (3)^3 - 3(3)^2 - 8(3) + 6

Step 2: Evaluate the function values

Now we will evaluate the function values calculated in the previous step: f(-1) = -1 - 3 - (-8) + 6 = 10 f(3) = 27 - 27 - 24 + 6 = -18

Step 3: Calculate the average rate of change

Next, we need to find the average rate of change, which is the change in function values divided by the change in x-values: Average rate of change = (f(3) - f(-1)) / (3 - (-1))

Step 4: Substitute the function values

Now substitute the function values found in Step 2 into the equation for the average rate of change: Average rate of change = (-18 - 10) / (3 - (-1))

Step 5: Calculate the result

Finally, we can calculate the result of the average rate of change: Average rate of change = (-28) / 4 = -7 The average rate of change of the function f(x) over the interval from x = -1 to x = 3 is -7.

Key Concepts

Polynomial Function

A polynomial function is a type of mathematical expression involving a sum of powers of variables with coefficients. These functions are incredibly versatile and arise frequently in numerous fields. The general form of a polynomial of degree \( n \) is:

• \( a_nx^n + a_{n-1}x^{n-1} + \, ..., + \, a_1x + a_0 \)

Here, \( a_n, a_{n-1}, \, ..., \, a_1, \, a_0 \) are constant coefficients, and \( n \) is a non-negative integer.

In the exercise at hand, the function \( f(x) = x^3 - 3x^2 - 8x + 6 \) is a cubic polynomial, as its highest power is 3. This polynomial describes a curve in the coordinate plane that can curve and twist in various manners, making it essential in modeling data that exhibit such trends.

Interval Evaluation

Interval evaluation involves assessing the function over a range of \( x \)-values to observe changes in the function's behavior. In our exercise, we evaluate the function \( f(x) = x^3 - 3x^2 - 8x + 6 \) over the interval from \( x = -1 \) to \( x = 3 \).

To perform this, calculate the function values at the boundaries of the interval. This means substituting \( x = -1 \) and \( x = 3 \) into the polynomial function:

• \( f(-1) = (-1)^3 - 3(-1)^2 - 8(-1) + 6 = 10 \)
• \( f(3) = (3)^3 - 3(3)^2 - 8(3) + 6 = -18 \)

By evaluating the polynomial for these specific values, you can see how the function behaves at each endpoint of the interval and prepare to determine the overall change across it.

Function Substitution

Function substitution is a crucial step in evaluating functions and solving various problems in mathematics. It involves replacing the variable \( x \) in the function with specific values to determine the function's output.

In our context, we substituted \( x = -1 \) and \( x = 3 \) into the given polynomial \( f(x) = x^3 - 3x^2 - 8x + 6 \). Here’s how substitution looks:

• When \( x = -1 \): Substitute \( -1 \) for \( x \) in the equation.
• When \( x = 3 \): Substitute \( 3 \) for \( x \) in the equation.

This process of substitution gives us the specific outputs 10 and -18 for these inputs, respectively. Substitution helps ascertain the function's precise output at given points and is not only a common practice in polynomial evaluation but also integral to understanding functions in calculus.

Calculus

Calculus is a branch of mathematics focusing on change and motion, utilizing concepts such as derivatives and integrals. An average rate of change is at its foundation. This concept shares similarities with the slope of a line and reflects how one quantity changes concerning another.

In this exercise, we computed the average rate of change of the function \( f(x) \) over the interval \( x = -1 \) to \( x = 3 \). The formula used is:

• \( \frac{f(b) - f(a)}{b - a} \)

Where \( a = -1 \) and \( b = 3 \).

By plugging in our previously derived function values, we found:

• \( \frac{-18 - 10}{3 - (-1)} = -7 \)

This result tells us that, on average, the function decreases by 7 units for every 1 unit increase in \( x \), across the specified interval. Understanding this concept will greatly aid in your comprehension of how rates of change form the basis for differentiation in calculus.