Functions of a Single Variable in Practice

Sep 14, 2024

Unit 2, Lesson 5: Writing Functions of a Single Variable

Overview

In this lesson, we focus on deriving functions of a single variable from word problems. The primary goal is to express desired outputs as functions of a single variable by manipulating inputs.

Key Objective

  • Learn to express real-world situations as mathematical functions.

Example 1: Pure Mathematics Example

Problem Statement

  • Situation: Let ( p ) be a point on the graph ( y = x^2 + 3 ).
  • Objective: Express the distance from the origin to the point ( p ) as a function of ( x ).

Strategy

  • Visualization: Draw a picture to help orient the problem.
    • Graph the parabola ( y = x^2 + 3 ).
    • Identify point ( P ) on the graph.

Solution

  • Distance Formula: Use the distance formula between two points:

    [ d = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} ]

  • Applying to Origin:

    • Origin ((0, 0)), and ( P(x, y) ).
    • Distance ( d = \sqrt{x^2 + y^2} ).
  • Function of ( x ):

    • Since ( y = x^2 + 3 ), substitute:

    [ d = \sqrt{x^2 + (x^2 + 3)^2} ]

  • Simplify:

    • Expand ((x^2 + 3)^2) to (x^4 + 6x^2 + 9).
    • Final function: ( f(x) = \sqrt{x^4 + 7x^2 + 9} ).

Example 2: Applied Problem - Fencing a Field

Problem Statement

  • Situation: A rectangular field is enclosed by a fence with an area of 500 square feet.
  • Objective: Express the feet of fencing needed to enclose the field as a function of the length of one side.

Strategy

  • Visualization: Draw the rectangle with length ( L ) and width ( W ).
  • Knowns:
    • Area = 500 square feet.
    • Perimeter = ( 2L + 2W ).

Solution

  • Transform Area Formula:

    • ( L \times W = 500 \rightarrow W = \frac{500}{L} ).
  • Perimeter as a Function of ( L ):

    • Substitute ( W ) in the perimeter formula:

    [ P = 2L + 2\left(\frac{500}{L}\right) ]

    • Simplify to: ( f(L) = 2L + \frac{1000}{L} ).
  • Function Representation:

    • Perimeter as a function of ( L ) or ( X ) (if ( X = L )):

    [ f(X) = 2X + \frac{1000}{X} ]

Conclusion

  • Practice problems available to reinforce these concepts.
  • Focus on understanding units and simplifying expressions to derive functional relationships from word problems.