Problem 15 - Applied Calculus

Question

Find a possible formula for the function represented by the data. $$ \begin{array}{c|c|c|c|c} \hline x & 0 & 1 & 2 & 3 \\ \hline f(x) & 4.30 & 6.02 & 8.43 & 11.80 \\ \hline \end{array} $$

Step-by-Step Solution

Answer

The function is \( f(x) = 0.345x^2 + 1.375x + 4.30 \).

Step 1: Identify the Pattern in Increments

Examine how the function values change when the variable increases. Look for differences between successive values:\( f(1) - f(0) = 6.02 - 4.30 = 1.72 \)\( f(2) - f(1) = 8.43 - 6.02 = 2.41 \)\( f(3) - f(2) = 11.80 - 8.43 = 3.37 \).

Step 2: Check for Linearity

Determine if the function might be linear by noting if differences between increments are constant. Since the differences are not constant, further investigation is needed to determine the type of function.

Step 3: Assume a Polynomial or Exponential Model

Try to see if the changes could be captured by a polynomial. Start by considering popular functions such as linear, quadratic, or exponential because they are commonly tested in such problems.

Step 4: Propose a Quadratic Model

A quadratic function takes the form \( f(x) = ax^2 + bx + c \). With three unknowns \( a, b, c \), we can create equations based on the given data points.

Step 5: Set Up Equations from Data Points

Using the function \( f(x) = ax^2 + bx + c \), substitute the given points:For \( x = 0 \), \( f(0) = c = 4.30 \).For \( x = 1 \), \( a(1)^2 + b(1) + 4.30 = 6.02 \) leading to \( a + b = 1.72 \).For \( x = 2 \), \( a(2)^2 + b(2) + 4.30 = 8.43 \) leading to \( 4a + 2b = 4.13 \).

Step 6: Solve the System of Equations

Using the equations \( a + b = 1.72 \) and \( 4a + 2b = 4.13 \), solve for \( a \) and \( b \):1. From \( a + b = 1.72 \) deduce \( b = 1.72 - a \).2. Substitute in \( 4a + 2b = 4.13 \), yielding:\( 4a + 2(1.72 - a) = 4.13 \) \( 4a + 3.44 - 2a = 4.13 \)\( 2a = 0.69 \)\( a = 0.345 \)Plug back to find \( b: b = 1.72 - 0.345 = 1.375 \).

Step 7: State the Function Formula

Use the calculated values \( a = 0.345 \), \( b = 1.375 \), and \( c = 4.30 \) to write the quadratic function:\( f(x) = 0.345x^2 + 1.375x + 4.30 \).

Key Concepts

Quadratic Functions

A quadratic function is one of the simplest types of polynomial functions. It comes in the form of:\[ f(x) = ax^2 + bx + c \]where \(a\), \(b\), and \(c\) are constants, and \(a\) should not be zero. This form characterizes what we call a parabola on a graph, which can open upwards (if \(a > 0\)) or downwards (if \(a < 0\)).
Quadratic functions are pivotal in calculus because they naturally model many physical phenomena, from the path of a ball thrown in the air to the shape of satellite antennas.
Some common properties to keep in mind about quadratic functions are:

• The vertex, which is the peak of the parabola. A precisely significant point.
• The axis of symmetry, a vertical line through the vertex.
• The roots or solutions where the function equals zero. Found using the quadratic formula.

Understanding these properties provides a strong foundation for solving problems involving quadratic functions.

Function Modeling

Function modeling involves writing equations to represent real-world phenomena. It's a bridge between theoretical mathematics and practical applications. When given a set of data, as in the exercise, identifying a suitable model can help make predictions or understand underlying trends.
In choosing potential models, quadratic functions are often considered because of their straightforward yet flexible nature. For our exercise, we delved into polynomials. This is because quadratic models, being simple yet expressive, are able to fit many sets of data accurately where linear functions fail.
Here's how you can begin function modeling:

• Start by plotting the data to visually identify the pattern and trend.
• Choose potential function types, based on how the data behaves, such as linear, quadratic, etc.
• Fit the chosen function to the data using algebraic methods to estimate the constants involved.

This process, while involving trial and error, grows more intuitive with practice.

Calculus Problem-Solving

Calculus is the study of change and motion, and it's often used to solve more complex problems that involve dynamic systems.
When dealing with functions in calculus, especially polynomial functions like quadratics, concepts such as derivatives and integrals come into play.
Here's how calculus can aid in problem-solving:

• **Derivatives** help identify rates of change, such as the speed at which a variable is changing within a function, and can be used to locate maxima or minima (think of the vertex in quadratic terms).
• **Integrals** are used to find accumulated quantities, like the area under a curve, which is crucial in physics and engineering.
• Understanding behavior and properties of functions can greatly enhance calculus solutions, making it an indispensable tool for engineers, scientists, and economists alike.

By understanding calculus thoroughly, you can effectively apply these tools to a broad spectrum of problems, making it easier to perform advanced function analysis.