Problem 16 - Algebra and Trigonometry

Question

A function is given. Determine the average rate of change of the function between the given values of the variable. $$ f(x)=4-x^{2} ; \quad x=1, x=1+h $$

Step-by-Step Solution

Answer

The average rate of change is \(-2 - h\).

Step 1: Understand the Formula for Average Rate of Change

The average rate of change of a function between two points is given by the formula: \( \frac{f(x_2) - f(x_1)}{x_2 - x_1} \), where \(x_1\) and \(x_2\) are the given values of the variable.

Step 2: Identify the Points

We have \(x_1 = 1\) and \(x_2 = 1 + h\). We need to find \(f(x_1)\) and \(f(x_2)\).

Step 3: Calculate \(f(x_1)\)

Substitute \(x_1 = 1\) into the function: \(f(1) = 4 - (1)^2 = 4 - 1 = 3\).

Step 4: Calculate \(f(x_2)\)

Substitute \(x_2 = 1 + h\) into the function: \(f(1+h) = 4 - (1+h)^2 = 4 - (1 + 2h + h^2) = 4 - 1 - 2h - h^2 = 3 - 2h - h^2\).

Step 5: Calculate the Difference in Function Values

Now, calculate \(f(x_2) - f(x_1)\): \((3 - 2h - h^2) - 3 = -2h - h^2\).

Step 6: Calculate the Difference in x Values

Find \(x_2 - x_1\): \((1 + h) - 1 = h\).

Step 7: Apply the Formula for the Average Rate of Change

Substitute into the average rate of change formula: \(\frac{-2h - h^2}{h}\). Simplify by dividing each term by \(h\): \(\frac{-2h}{h} - \frac{h^2}{h} = -2 - h\).

Key Concepts

Quadratic Functions

A quadratic function is a type of polynomial function that takes the form \( f(x) = ax^2 + bx + c \). It is called quadratic because the highest degree term is squared. In our example, the function given is \( f(x) = 4 - x^2 \), which can be rewritten as \( -x^2 + 0x + 4 \). This shows the standard form of a quadratic function where:

• \( a = -1 \)
• \( b = 0 \)
• \( c = 4 \)

Quadratic functions have a symmetrical graph called a parabola. For our function, since the \( a \) value is negative, the parabola opens downwards. The general approach of working with quadratic functions involves identifying such coefficients, understanding the direction of the opening, and analyzing how the function behaves throughout its domain. By understanding these characteristics, evaluating the function across different values becomes more intuitive. Evaluating the average rate of change for a quadratic function can initially appear daunting, but breaking it down into manageable steps, as we will see, simplifies the process.

Function Evaluation

Function evaluation refers to the process of substituting a particular value of \( x \) into the function to find the corresponding \( f(x) \). For a given function \( f(x) = 4 - x^2 \), we evaluate the function at specific values of \( x \) to find the outputs. For example:

• At \( x = 1 \), substitute into the function to find \( f(1) = 4 - (1)^2 = 3 \).
• At \( x = 1 + h \), substitute to find \( f(1+h) = 4 - (1 + h)^2 = 3 - 2h - h^2 \).

These calculations demonstrate how the function behavior can change depending on the inputs you select. When evaluating functions, particularly quadratic ones, consider the order of operations: square the value of \( x \), then multiply by any coefficients before applying addition or subtraction. This ensures an accurate result at each step. Function evaluation is a fundamental skill in mathematics and helps in understanding how functions react to changes in their input.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions to make them easier to work with or to solve a particular problem. In our case, we need to find the average rate of change between the points \( x = 1 \) and \( x = 1+h \). Here is how algebra helps in simplifying the calculations: 1. Start with function outputs: identify what needs to be simplified, \( f(1) = 3 \) and \( f(1+h) = 3 - 2h - h^2 \).2. Difference calculation: \( f(1+h) - f(1) = (3 - 2h - h^2) - 3 = -2h - h^2 \).3. Difference in \( x \)-values: \( x_2 - x_1 = (1 + h) - 1 = h \).Once we have these parts, we compute the average rate of change: \[ \frac{-2h - h^2}{h} = -2 - h \]By executing each step carefully, using proper arithmetic and algebra, results in a simplified expression that represents the change in the function's value concerning changes in \( x \). Algebraic manipulation turns complex expressions into simpler, solvable forms. It is a critical skill not only in simplifying equations but also in efficiently solving mathematical problems.