Module 8 - Continuity
  Introduction | Lesson 1 | Lesson 2 | Lesson 3 | Self-Test
 Lesson 8.3: Piecewise Functions

In this lesson you will modify a piecewise function to make it continuous and then graph the function using Boolean operators.

Consider the function defined by

8.3.1 By using the definition of continuity, find the value of k that makes the function continuous at
x = 2. Click here for the answer.

Graphing a Piecewise Function

Display the graph of using the value of k that makes the function continuous.

  • Set Xres = 1.
  • Enter
    You can paste " " from the Test menu by pressing [TEST] . [TEST] is above .
  • Enter Y2 = ( (1/2)X2+3 ) / (X>2)
    The symbol ">" is in the TEST menu.
  • Graph the function in a [-3, 7, 1] x [-3, 25, 5] window.

The graph appears to be continuous at x = 2.

8.3.2 Describe how the graph would change if you changed k to 7. Click here for the answer.

Illustrate the result of changing the value of k to 7.

  • Set Y2=( (1/2)X2+7 ) / (X>2).

Editing Existing Expressions

Rather than retyping the entire expression in Y2 you can simply replace the "3" with a "7". Move the cursor on to the "3" and press . The old value should be replaced since the calculator is in the overwrite mode.

The calculator shows it is in the overwrite mode when the cursor is a blinking solid rectangle. It can be changed from overwrite to insert mode and back again by pressing [INS]. ([INS] is above in the first row of keys.) Each time you press [INS] the mode changes, or "toggles," between overwrite and insert. In insert mode, the cursor is a blinking underline character. Whenever an arrow key or is pressed, the calculator returns to overwrite mode.

  • Display the graph of the function by pressing .

The second part of the piecewise function has been shifted upward, and the function is no longer continuous at x = 2. The discontinuity in this modified function (k = 7) is a jump discontinuity.

8.3.3 Evaluate the left- and right-hand limits of the piecewise function with k = 7 at x = 2. Click here for the answer.

< Back | Next >
  ©Copyright 2007 All rights reserved. | Trademarks | Privacy Policy | Link Policy