Problem 11 - Contemporary Precalculus

Question

Find the average rate of change of the function f over the given interval. $$f(x)=3+x^{3} \text { from } x=0 \text { to } x=2$$

Step-by-Step Solution

Answer

Answer: The average rate of change is $$4$$.

Step 1: Plug in the values for x in the function

First, we need to find the values for $$f(0)$$ and $$f(2)$$ by plugging the values of $$x=0$$ and $$x=2$$ into the given function $$f(x) = 3 + x^3$$: $$f(0) = 3 + (0)^3 = 3$$ $$f(2) = 3 + (2)^3 = 3 + 8 = 11$$

Step 2: Apply the formula for the average rate of change

Now that we have the values for $$f(0)$$ and $$f(2)$$, we can use the formula for the average rate of change: $$\text{Average rate of change = }\frac{f(b)-f(a)}{b-a}$$ In this case, $$a=0$$ and $$b=2$$.

Step 3: Calculate the average rate of change

Plugging the values of $$f(0), f(2), a,$$ and $$b$$ into the formula, we find the average rate of change as: $$\text{Average rate of change }= \frac{f(2)-f(0)}{2-0} = \frac{11-3}{2} = \frac{8}{2}$$ Thus, the average rate of change is $$4$$.

Key Concepts

Precalculus

Precalculus is a course that prepares students for the study of calculus. It revises and builds upon concepts from algebra and geometry to establish the foundational skills needed for calculus. Here, concepts like functions, their properties, and transformations are explored in depth. Understanding how a function behaves and changes forms a big part of precalculus.

Functions represent relationships between variables—how one quantity changes in response to another. In the case of our exercise, the function described by \( f(x) = 3 + x^3 \) maps any number \( x \) to a value that is three more than the cube of \( x \). By investigating the behavior of this function over a certain interval, we lay the groundwork for the more complex concepts of calculus that follow, such as limits, derivatives, and integrals.

Rate of Change

The concept of 'rate of change' refers to how much a quantity changes with respect to changes in another quantity. In our context, this is the change in the function's output value over a particular range of input values. It's the precalculus equivalent of the derivative in calculus, which describes instantaneous rate of change; however, average rate of change means we're looking at the change over a specific interval.

To find the average rate of change for the function \( f(x) = 3 + x^3 \) from \( x = 0 \) to \( x = 2 \), we calculate the difference in function values at the endpoints of the interval and divide by the difference in the \( x \) values. It's essentially the slope of the secant line that runs through the points on the graph at those intervals. A positive rate of change signifies an increase, while a negative one indicates a decrease in function value as the input increases.

Polynomial Functions

Polynomial functions, such as the one in our exercise \( f(x) = 3 + x^3 \), are characterized by having terms with non-negative integer exponents. These functions are fundamental in mathematical analysis and appear frequently in various contexts. A polynomial's degree is determined by the highest power of \( x \) present in the function; for our function, the degree is three, making it a cubic polynomial.

Cubic polynomials like this one tend to start low, bend to a peak or trough, and then bend back to increase or decrease indefinitely. The average rate of change between two points on a polynomial can give you a sense of how steeply the function rises or falls on that interval. Recognizing the shape and features of the graph of a polynomial aids in understanding its rate of change and anticipating its behavior at different intervals.