Name: _________________________________________________ Period: ___________ Date: ________________ Continuity, End Behavior, and Limits Guided Notes The graph of a continuous function has no breaks, holes, or gaps. You can trace the graph of a continuous function without lifting your pencil. One condition for a function π(π) to be continuous at π = π is that the function must approach a unique function value as π -values approach π from the left and right sides. The concept of approaching a value without necessarily ever reaching it is called a limit. If the value of π(π)approaches a unique value π³ as π approaches π from each side, then the limit of π(π) as π approaches π is π³. π₯π’π¦ π(π) = π³ π→π Functions that are not continuous are discontinuous. Graphs that are discontinuous can exhibit: ο· Infinite discontinuity (A function has an infinite discontinuity at π = π if the function value increases or decreases indefinitely as π approaches π from the left and right) ο· Jump discontinuity,( A function has a jump discontinuity at π = π if the limits of the function as π approaches π from the left and right exist but have two distinct values. ο· Removable discontinuity, also called point discontinuity (function has a removable discontinuity if the function is continuous everywhere except for a hole at π = π. Continuity Test A function π(π) is continuous at π = π if it satisfies the following conditions. 1. π(π) is defined at π. π(π)exists. 2. π(π) approaches the same function value to the left and right of π. π₯π’π¦ π(π) ππππππ 3. The function value that π(π)approaches from each side of π is π(π). π₯π’π¦ π(π) = π (π) π→π π→π Sample Problem 1: Determine whether each function is continuous at the given π -values. Justify using the continuity test. If discontinuous, identify the type of discontinuity. a. π(π) = πππ + π − π ππ π = π π(π) = πππ + π − π ππ π = π π π(π) = π ∗ π + π − π π(π) = −π π(π) ππππππ π → π− π → −π y 30 20 10 x -40 -30 -20 -10 10 20 30 π π. π π. ππ π. πππ π(π) −π. ππ −π. ππππ −π. ππππππ π π. π π. ππ π. πππ π(π) −π. ππ −π. ππππ −π. ππππππ π → π+ π → −π 40 -10 -20 -30 π(π) = −π πππ π → −π ππππ ππππ πππ π ππ π = π π₯π’π¦ πππ + π − π = π (π) π→π π(π) = πππ + π − π ππ ππππππππππ ππ π = π Copyright © PreCalculusCoach.com 1 Name: _________________________________________________ Period: ___________ Date: ________________ Continuity, End Behavior, and Limits Guided Notes b. π(π) = |ππ| π |ππ| ππ π = π π |π ∗ π| π π(π) = = π π π»ππ ππππππππ ππ πππ ππππππ ππ π = π π → π− π → −π ππ π = π 5 π(π) = y 4 3 2 1 x -5 -4 -3 -2 -1 1 2 3 4 5 π −π. π −π. ππ −π. πππ π(π) −π −π −π π π. π π. ππ π. πππ π(π) π π π π → π+ π → π -1 -2 -3 -4 |ππ| πππ ππππ π ππππππππππππ ππ π = π π πππππ π ππππππ πππ π πππ − π ππ ππππππππ πππ ππ ππ π = π -5 c. π(π) = ππ − π π+π π(π) = ππ − π ππ π = −π π+π (−π)π − π π π(−π) = = −π + π π π(π) ππ πππ ππππππ ππ π = −π ππ π = −π 5 π(π) = y 4 ππ −π π(π) = π+π ππ π ππππππππππππ ππ π = −π π → −π− π → −π 3 2 π −π. π −π. ππ −π. πππ π(π) −π. π −π. ππ −π. πππ 1 x -5 -4 -3 -2 -1 1 -1 2 3 4 5 π → −π+ π → −π -2 π −π. π −π. ππ −π. πππ -3 π(π) −π. π −π. ππ −π. πππ -4 -5 ππ − π πππ πππππ π ππππππππππππ ππ π = −π π+π πππππ π πππππ ππ − π ππ ππππππππ πππ ππ ππ π = −π π(π) = Copyright © PreCalculusCoach.com 2 Name: _________________________________________________ Period: ___________ Date: ________________ Continuity, End Behavior, and Limits Guided Notes d. π(π) = π πππ π ππ π = π πππ π π(π) = = ∞ π ∗ ππ π(π) ππ πππ ππππππ ππ π = π π π(π) = π ππ π ππππππππππππ ππ π = π ππ π → π− π → +∞ ππ π = π 5 π(π) = y 4 3 2 1 x -5 -4 -3 -2 -1 1 2 3 4 5 -1 π −π. π −π. ππ −π. πππ π(π) ππ. ππ π, πππ. ππ πππ, πππ. ππ π π. π π. ππ π. πππ π(π) ππ. ππ π, πππ. ππ πππ, πππ. ππ π → π+ π → +∞ -2 -3 -4 π πππ ππππππππ π ππππππππππππ ππ π = π πππππ πππ π πππππ ππ + ∞ ππππ π → π -5 π(π) = Intermediate Value Theorem If π(π)is a continuous function and π < π and there is a value π such that π is between π(π) and π(π) , then there is a numberπ, such that π < π < π and π(π) = π The Location Principle If π(π) is a continuous function and π(π) and π(π) have opposite signs, then there exists at least one value π, such that π < π < π and π(π) = π. That is, there is a zero between π and π. Sample Problem 2: Determine between which consecutive integers the real zeros of function are located on the given interval. a. π(π) = (π − π)π − π 5 [π, π] y 4 π π π π π π π π π π π −π −π −π π π 3 2 1 x -5 -4 -3 -2 -1 1 2 -1 3 4 5 π(π) is positive and π(π) is negative, π(π) ππππππ ππππ ππ π ≤ π ≤ π π(π) is negative and π(π) is positive π(π) ππππππ ππππ ππ π ≤ π ≤ π -2 -3 -4 π(π) has zeros in intervals: π ≤ π ≤ π πππ π ≤ π ≤ π -5 Copyright © PreCalculusCoach.com 3 Name: _________________________________________________ Period: ___________ Date: ________________ Continuity, End Behavior, and Limits Guided Notes End Behavior The end behavior of a function describes what the π -values do as |π| becomes greater and greater. When π becomes greater and greater, we say that π approaches infinity, and we write π → +∞. When π becomes more and more negative, we say that π approaches negative infinity, and we write π → −∞. The same notation can also be used with π or π(π) and with real numbers instead of infinity. Left - End Behavior (as π becomes more and more negative): π₯π’π¦ π(π) π→−∞ Right - End Behavior (as π becomes more and more positive): π₯π’π¦ π(π) π→+∞ The π(π) values may approach negative infinity, positive infinity, or a specific value. Sample Problem 3: Use the graph of each function to describe its end behavior. Support the conjecture numerically. a. π(π) = −πππ + ππ − π From the graph, it appears that: π(π) → ∞ as π → −∞ and π(π) → −∞ – as π → ∞ y 8 7 6 The table supports this conjecture. 5 4 3 π 2 1 x -9 -8 -7 -6 -5 -4 -3 -2 -1-1 1 2 3 4 5 6 7 8 −πππ π π ∗ ππππ −πππ π πππ π ∗ πππ −π -π ∗ πππ πππ -π ∗ ππππ 9 -2 -3 -4 -5 -6 -7 -8 b. π(π) = ππ − π π−π From the graph, it appears that: π(π) → −π as π → −∞ and π(π) → −π as π → ∞ y 12 10 The table supports this conjecture. 8 6 4 π 2 -14 -12 -10 -8 -6 -4 -2 x 2 4 6 8 10 12 −πππ π −π. ππππ −π. ππππ 14 -2 -4 -6 -8 -10 -12 Copyright © PreCalculusCoach.com −πππ 4 π πππ πππ −π. ππ −π. πππ −π. ππππ Name: _________________________________________________ Period: ___________ Date: ________________ Continuity, End Behavior, and Limits Guided Notes Increasing, Decreasing, and Constant Functions A function π is increasing on an interval π° if and only if for every π and π contained in π°, (π) < π(π) , whenever π < π . A function π is decreasing on an interval π° if and only if for every π and π contained in π°, π(π) > π(π) whenever π < π . A function π remains constant on an interval π° if and only if for every π and π contained in π°, π(π) = π(π) whenever π<π . Points in the domain of a function where the function changes from increasing to decreasing or from decreasing to increasing are called critical points. Sample Problem 4: Determine the interval(s) on which the function is increasing and the interval(s) on which the function is decreasing. a. π(π) = ππ − ππ + π 5 From the graph, it appears that: A function ππ − ππ + π is decreasing for π < π. π A function ππ − ππ + π is increasing for π > π. π y 4 3 2 The table supports this conjecture. 1 x -5 -4 -3 -2 -1 1 2 3 4 5 -1 -2 π −π π π π π π π π. π π −π. ππ −π. π π π -3 -4 -5 b. π π From the graph, it appears that: π A function ππ − π ππ − πππ + π is increasing: π < −π. ππ πππ π > π π A function ππ − ππ − πππ + π is decreasing: π −π. ππ < π < π π(π) = ππ − ππ − πππ + π y 12 10 8 6 The table supports this conjecture. 4 2 -14 -12 -10 -8 -6 -4 -2 x 2 4 6 8 10 12 14 π −π −π. ππ −π π π π π ππ ππ. ππ ππ. π π −ππ −π. π -2 -4 -6 -8 -10 -12 Copyright © PreCalculusCoach.com 5
