Problem 16 - Precalculus Essentials

Question

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

Step-by-Step Solution

Answer

The average rate of change of the function \(f(x)=x^2-2x\) from \(x_1=3\) to \(x_2=6\) is 7.

Step 1: Compute \(f(x_1)\)

Substitute \(x_1=3\) in the expression \(f(x)=x^2-2x\) to obtain \(f(x_1)\). The calculation is as follow: \(f(x_1)= f(3) = 3^2-2*3 = 9-6 = 3\)

Step 2: Compute \(f(x_2)\)

Substitute \(x_2=6\) in the expression \(f(x)=x^2-2x\) to obtain \(f(x_2)\). The calculation is as follow: \(f(x_2)=f(6)=6^2-2*6=36-12=24\)

Step 3: Compute the Average Rate of Change

The average rate of change is given by the formula \(\frac{f(x_2)-f(x_1)}{x_2-x_1}\). Substituting \(f(x_1)=3\), \(f(x_2)=24\), \(x_1=3\), and \(x_2=6\), we get: \(\frac{24-3}{6-3} = \frac{21}{3} = 7\).

Key Concepts

Function Evaluation

Function evaluation is a fundamental concept in algebra and calculus. It involves substituting specific values into a function to discover the output. In the context of our problem, we need to evaluate the quadratic function \( f(x) = x^2 - 2x \) at particular points. These specific points are \( x_1 = 3 \) and \( x_2 = 6 \).
To evaluate at these points:

• For \( x_1 = 3 \): Substitute 3 into the function, i.e., \( f(3) = 3^2 - 2 \times 3 \). This gives us the result \( f(3) = 3 \).
• For \( x_2 = 6 \): Similarly, substitute 6 into the function, i.e., \( f(6) = 6^2 - 2 \times 6 \), which results in \( f(6) = 24 \).

This process of plugging in values helps us calculate the function's output at these specific inputs, revealing critical information about the behavior of the function over a particular range.

Quadratic Functions

Quadratic functions are polynomial expressions of the form \( ax^2 + bx + c \) where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). The characteristic U-shaped graph of a quadratic function is known as a parabola.
In our problem, the function is \( f(x) = x^2 - 2x \):

• The term \( x^2 \) is the quadratic term, giving the function its parabolic shape.
• The term \( -2x \) is the linear term, which shifts the parabola left or right.
• No constant term is present, starting the parabola at the origin and shifting according to the linear term.

This function graphically represents how the values change as \( x \) increases. The unique properties of quadratics allow them to model various physical phenomena, such as projectile motion, optimization problems, and more, which is why understanding them is so crucial.

Difference Quotient

The difference quotient is a crucial concept in calculus and serves as the foundation for determining the rate of change for functions. It is instrumental in understanding how we transition from algebraic expressions to differential calculus.
The difference quotient formula is expressed as\[\frac{f(x_2) - f(x_1)}{x_2 - x_1}\]This formula calculates the average rate of change of the function between two points, \( x_1 \) and \( x_2 \). In our example:

• The values \( x_1 = 3 \) and \( x_2 = 6 \) are substituted into the function \( f(x) = x^2 - 2x \).
• We calculated \( f(x_1) = 3 \) and \( f(x_2) = 24 \).

Substituting these into the difference quotient formula gives:\[\frac{24 - 3}{6 - 3} = \frac{21}{3} = 7\]This result, 7, represents the average rate of change of the function over that interval. Understanding the difference quotient can simplify complex functions into insightful, understandable rates at which changes occur within a specified interval.