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MR. JOJO L. CALAGUE MATHEMATICS TEACHER
CONTINUITY OF A FUNCTION c c c c Graph 1 Graph 2 Graph 3 Graph 4
CONTINUITY OF A FUNCTION c Graph 1 Graph 2 c οƒΌ The function is not defined at c οƒΌ The limit of f as x β†’ 𝒄 exist οƒΌ Both the value of the function at c and the limit as x β†’ 𝒄 exist but not equal
CONTINUITY OF A FUNCTION c c Graph 3 Graph 4 οƒΌ The limit of f as x approaches c does not exist CONTINUOUS FUNCTION οƒΌ The value of the function is defined, that is f(c) οƒΌ The limit of f exist, that is f(c) οƒΌ The value of f and limit of f are equal
CONTINUITY AT A POINT Definition A function f is continuous at c if and only if lim π‘₯→𝑐 𝑓 π‘₯ = 𝑓(𝑐) This implies that the three conditions must be satisfied: (1) f(c) exist; (2) lim π‘₯→𝑐 𝑓 π‘₯ exist; an d (3) lim π‘₯→𝑐 𝑓 π‘₯ = f(c)
CONTINUITY AT A POINT Types of Discontinuity Type 1 Removable Discontinuit y 𝐼𝑓 𝑓 π‘ π‘Žπ‘‘π‘–π‘ π‘“π‘–π‘’π‘  π‘π‘œπ‘›π‘‘π‘–π‘‘π‘–π‘œπ‘› 2 , 𝑏𝑒𝑑 π‘“π‘Žπ‘–π‘™π‘  π‘‘π‘œ π‘π‘œπ‘›π‘‘π‘–π‘‘π‘–π‘œπ‘›π‘  1 π‘œπ‘Ÿ 3 , π‘‘β„Žπ‘’ π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘–π‘‘π‘¦ 𝑖𝑠 π‘π‘Žπ‘™π‘™π‘’π‘‘ π‘Ÿπ‘’π‘šπ‘œπ‘£π‘Žπ‘π‘™π‘’. π‘‡β„Žπ‘–π‘  π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘–π‘‘π‘¦ π‘œπ‘π‘π‘’π‘Ÿπ‘  π‘€β„Žπ‘’π‘› π‘‘β„Žπ‘’π‘Ÿπ‘’ 𝑖𝑠 π‘Ž β„Žπ‘œπ‘™π‘’ 𝑖𝑛 π‘‘β„Žπ‘’ π‘”π‘Ÿπ‘Žπ‘β„Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘›
CONTINUITY AT A POINT Types of Discontinuity Type 2 Jump or Essential Discontinui ty 𝐼𝑓 𝑓 𝑖𝑠 π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘  π‘Žπ‘‘ 𝑐 π‘Žπ‘›π‘‘ π‘π‘œπ‘›π‘‘π‘–π‘‘π‘–π‘œπ‘› 2 𝑖𝑠 π‘›π‘œπ‘‘ π‘ π‘Žπ‘‘π‘–π‘ π‘“π‘–π‘’π‘‘, π‘‘β„Žπ‘’ π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘–π‘‘π‘¦ 𝑖𝑠 π‘π‘Žπ‘™π‘™π‘’π‘‘ π‘’π‘ π‘ π‘’π‘›π‘‘π‘–π‘Žπ‘™. π‘‡β„Žπ‘–π‘  π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘–π‘‘π‘¦ π‘œπ‘π‘π‘’π‘Ÿπ‘  π‘€β„Žπ‘’π‘› π‘‘β„Žπ‘’π‘”π‘Ÿπ‘Žπ‘β„Ž π‘ π‘‘π‘œπ‘π‘  π‘Žπ‘‘ π‘œπ‘›π‘’ π‘π‘œπ‘–π‘›π‘‘ π‘Žπ‘›π‘‘ π‘ π‘’π‘’π‘šπ‘‘ π‘‘π‘œ π‘—π‘’π‘šπ‘ π‘Žπ‘‘ π‘Žπ‘› π‘œπ‘‘β„Žπ‘’π‘Ÿ π‘π‘œπ‘–π‘›π‘‘. π‘‡β„Žπ‘’ 𝑙𝑒𝑓𝑑 β„Žπ‘Žπ‘›π‘‘ π‘Žπ‘›π‘‘ π‘‘β„Žπ‘’ π‘Ÿπ‘–π‘”β„Žπ‘‘ β„Žπ‘Žπ‘›π‘‘ π‘™π‘–π‘šπ‘–π‘‘π‘  𝑒π‘₯𝑖𝑠𝑑 𝑏𝑒𝑑 π‘Žπ‘Ÿπ‘’ π‘›π‘œπ‘‘ π‘’π‘žπ‘’π‘Žπ‘™.
CONTINUITY AT A POINT Types of Discontinuity Type 3 Asymptotic or Infinite Discontinuity 𝐼𝑓 𝑓 𝑖𝑠 π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘  π‘Žπ‘‘ 𝑐 π‘Žπ‘›π‘‘ π‘π‘œπ‘›π‘‘π‘–π‘‘π‘–π‘œπ‘› 2 𝑖𝑠 π‘›π‘œπ‘‘ π‘ π‘Žπ‘‘π‘–π‘ π‘“π‘–π‘’π‘‘ π‘Žπ‘›π‘‘ π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ π‘™π‘–π‘šπ‘–π‘‘ π‘œπ‘“ π‘‘β„Žπ‘’ π‘‘π‘€π‘œ π‘™π‘–π‘šπ‘–π‘‘π‘  𝑖𝑠 𝑖𝑛𝑓𝑖𝑛𝑖𝑑𝑒 π‘‘β„Žπ‘’ π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘–π‘‘π‘¦ 𝑖𝑠 π‘π‘Žπ‘™π‘™π‘’π‘‘ 𝑖𝑛𝑓𝑖𝑛𝑖𝑑𝑒.
EXAMPLE PRACTICE YOUR SKILLS! Given the graph of f(x),shown bel ow, determine if f(x) is continu ous at (1) x = -2 (2) x = 0 (3) x = 3
EXAMPLE PRACTICE YOUR SKILLS! Given the graph of f(x),shown belo w, determine if f(x) is continuous at x = -2. (1) f(-2) =2 (2) lim π‘₯β†’βˆ’2 𝑓 π‘₯ does not exi st (3) lim π‘₯β†’βˆ’2 𝑓 π‘₯ β‰  f(-2) f(x) is not continuous at x = -2
EXAMPLE PRACTICE YOUR SKILLS! Given the graph of f(x),shown belo w, determine if f(x) is continuous at x = 0. (1) f(0) =1 (2) lim π‘₯β†’0 𝑓 π‘₯ = 1, limit exis t (3) lim π‘₯β†’0 𝑓 π‘₯ = f(0) f(x) is continuous at x = 0
EXAMPLE PRACTICE YOUR SKILLS! Given the graph of f(x),shown belo w, determine if f(x) is continuous at x = 3. (1) f(3) =-1 (2) lim π‘₯β†’3 𝑓 π‘₯ does not ex ist (3) lim π‘₯β†’3 𝑓 π‘₯ β‰  f(3) f(x) is not continuous at x = 3
Determine whether or not the following ar e continuous functions. a. f(x) = π’™πŸ βˆ’π’™ βˆ’πŸπŸ 𝒙 βˆ’πŸ’ at x = 4 b. 𝒇 𝒙 = 𝒙 + 𝟏, π’Šπ’‡ 𝒙 < πŸ’ 𝒙 βˆ’ πŸ’ 𝟐 + πŸ‘ π’Šπ’‡ 𝒙 β‰₯ πŸ’ at x = 1 c. 𝒇 𝒙 = 𝟏 𝒙 at x = 0 EXAMPLE
Solution f(x) = π’™πŸ βˆ’π’™βˆ’πŸπŸ 𝒙 βˆ’πŸ’ x = 4 f(4) = (πŸ’)𝟐 βˆ’(πŸ’)+𝟏𝟐 πŸ’ βˆ’πŸ’ f(4) = π’–π’π’…π’†π’‡π’Šπ’π’†π’… Condition 1
Solution f(x) = π’™πŸ βˆ’π’™βˆ’πŸπŸ 𝒙 βˆ’πŸ’ x = 4 lim π‘₯β†’0 π’™πŸ βˆ’π’™ βˆ’ 𝟏𝟐 𝒙 βˆ’ πŸ’ Condition 2 = lim π‘₯β†’0 𝒙 βˆ’πŸ’ (𝒙+πŸ‘) 𝒙 βˆ’πŸ’ = lim π‘₯β†’0 (𝒙 + πŸ‘) = 7
Solution f(x) = π’™πŸ βˆ’π’™βˆ’πŸπŸ 𝒙 βˆ’πŸ’ x = 4 Condition 3 lim π‘₯β†’4 𝑓 π‘₯ β‰  f(4)
Solution f(x) = π’™πŸ βˆ’π’™βˆ’πŸπŸ 𝒙 βˆ’πŸ’ x = 4 𝐼𝑓 𝑓 π‘ π‘Žπ‘‘π‘–π‘ π‘“π‘–π‘’π‘  π‘π‘œπ‘›π‘‘π‘–π‘‘π‘–π‘œπ‘› 2 , 𝑏𝑒𝑑 π‘“π‘Žπ‘–π‘™π‘  π‘‘π‘œ π‘π‘œπ‘›π‘‘π‘–π‘‘π‘–π‘œπ‘›π‘  1 π‘œπ‘Ÿ 3 , π‘‘β„Žπ‘’ π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘–π‘‘π‘¦ 𝑖𝑠 π‘π‘Žπ‘™π‘™π‘’π‘‘ π‘Ÿπ‘’π‘šπ‘œπ‘£π‘Žπ‘π‘™π‘’.
Solution f(x) = π’™πŸ βˆ’π’™βˆ’πŸπŸ 𝒙 βˆ’πŸ’ x = 4 𝐻𝑒𝑛𝑐𝑒, 𝑓 π‘₯ 𝑖𝑠 π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘œπ‘’π‘  π‘Žπ‘‘ π‘₯ = 4
Solution 𝒇 𝒙 = πŸ•, 𝒙 = πŸ’ π’™πŸ βˆ’π’™βˆ’πŸπŸ 𝒙 βˆ’πŸ’ , 𝒙 β‰  πŸ’ 𝑓 π‘₯ 𝑖𝑠 π‘π‘œπ‘›π‘‘π‘–π‘›π‘œπ‘’π‘  π‘Žπ‘‘ π‘₯ = 4 Redefining the function to make it a coπ‘›π‘‘π‘–π‘›π‘œπ‘’π‘  π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘›
Solution 𝒇 𝒙 = 𝒙 + 𝟏, π’Šπ’‡ 𝒙 < πŸ’ 𝒙 βˆ’ πŸ’ 𝟐 + πŸ‘ π’Šπ’‡ 𝒙 β‰₯ πŸ’ x = 4 f(4) = πŸ’ βˆ’ πŸ’ 𝟐 + πŸ‘ f(4) = πŸ‘ Condition 1
Solution lim π‘₯β†’4 𝒇(𝒙) Condition 2 = 𝐷𝑁𝐸 𝒇 𝒙 = 𝒙 + 𝟏, π’Šπ’‡ 𝒙 < πŸ’ 𝒙 βˆ’ πŸ’ 𝟐 + πŸ‘ π’Šπ’‡ 𝒙 β‰₯ πŸ’ x = 4 𝐼𝑓 𝑓 𝑖𝑠 π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘  π‘Žπ‘‘ 𝑐 π‘Žπ‘›π‘‘ π‘π‘œπ‘›π‘‘π‘–π‘‘π‘–π‘œπ‘› 2 𝑖𝑠 π‘›π‘œπ‘‘ π‘ π‘Žπ‘‘π‘–π‘ π‘“π‘–π‘’π‘‘, π‘‘β„Žπ‘’ π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘–π‘‘π‘¦ 𝑖𝑠 π‘π‘Žπ‘™π‘™π‘’π‘‘ π‘’π‘ π‘ π‘’π‘›π‘‘π‘–π‘Žπ‘™. lim π‘₯β†’4βˆ’ 𝒇 𝒙 = 5 lim π‘₯β†’4+ 𝒇 𝒙 = 3
Solution Condition 3 𝒇 𝒙 = 𝒙 + 𝟏, π’Šπ’‡ 𝒙 < πŸ’ 𝒙 βˆ’ πŸ’ 𝟐 + πŸ‘ π’Šπ’‡ 𝒙 β‰₯ πŸ’ x = 4 lim π‘₯β†’4 𝑓 π‘₯ β‰  f(4) 𝐻𝑒𝑛𝑐𝑒, 𝑓 π‘₯ 𝑖𝑠 π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘œπ‘’π‘  π‘Žπ‘‘ π‘₯ = 4
Solution 𝒇 𝒙 = 𝟏 𝒙 𝒙 = 𝟎 f(0) = 𝟏 𝒙 f(0) = π’–π’π’…π’†π’‡π’Šπ’π’†π’… Condition 1
Solution 𝒇 𝒙 = 𝟏 𝒙 𝒙 = 𝟎 Condition 2 lim π‘₯β†’0+ 𝒇 𝒙 = +∞ lim π‘₯β†’0βˆ’ 𝒇 𝒙 = βˆ’βˆž
Solution 𝒇 𝒙 = 𝟏 𝒙 𝒙 = 𝟎 𝐻𝑒𝑛𝑐𝑒, 𝑓 π‘₯ 𝑖𝑠 π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘œπ‘’π‘  π‘Žπ‘‘ π‘₯ = 0 and f(x) is said to have an infinite discontinuity at x = 0.
Exercises 1. f(x) = π’™πŸ βˆ’πŸ— 𝒙 βˆ’πŸ‘ at x = 2 2. 𝒇 𝒙 = βˆ’πŸπ’™ + πŸ’, π’Šπ’‡ 𝒙 β‰₯ πŸ‘ 𝒙 βˆ’ 𝟏 π’Šπ’‡ 𝒙 < πŸ‘ at x = 3 4. 𝒇 𝒙 = π’™πŸβˆ’πŸ 𝒙 βˆ’πŸ at x = 1 3. 𝒇 𝒙 = πŸπ’™ βˆ’πŸ‘ 𝒙+ πŸ‘ at x = -3 π·π‘’π‘‘π‘’π‘Ÿπ‘šπ‘–π‘›π‘’ 𝑖𝑓 π‘‘β„Žπ‘’ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘› 𝑖𝑠 π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘  π‘Žπ‘‘ 𝑔𝑖𝑣𝑒𝑛 π‘›π‘’π‘šπ‘π‘’π‘Ÿ 𝑐.
CONTINUITY ON AN INTERVAL Definition A function f is continuous on an open interval (a, b) if f is continuous at each number in the interv al A function f is continuous on a closed inte rval [a, b] if (𝐛) lim π‘₯β†’π‘Ž+ 𝑓 π‘₯ = f(a) and lim π‘₯β†’π‘βˆ’ 𝑓 π‘₯ = f(b) (a) f is continuous on the open interval (a, b)
Determine if the function is continuous on the indicated closed interval. a. f(x) = 𝒙 𝒙 ; [𝟎, 𝟏] b. 𝒇 𝒙 = πŸ— βˆ’ π’™πŸ ; [-3, 3] c. 𝒇 𝒙 = 𝟏 π’™πŸ βˆ’πŸ ; [0,1] EXAMPLE
Solution: f(x) = 𝒙 𝒙 ; [𝟎, 𝟏] a. f(a) = ? f(x) = 𝒙 𝒙 f(0) = 𝟎 𝟎 f(0) = 𝟎 f(b) = ? f(x) = 𝒙 𝒙 f(1) = 𝟏 𝟏 f(1) = 𝟏 b. lim π‘₯β†’0+ 𝑓 π‘₯ and lim π‘₯β†’1βˆ’ 𝑓 π‘₯ lim π‘₯β†’0+ 𝒙 𝒙 = 𝟎 𝟎 lim π‘₯β†’0+ 𝒙 𝒙 = 0 lim π‘₯β†’1βˆ’ 𝒙 𝒙 = 𝟏 𝟏 lim π‘₯β†’1βˆ’ 𝒙 𝒙 = 1
Solution: f(x) = 𝒙 𝒙 ; [𝟎, 𝟏] f(0) = 𝟎 f(1) = 𝟏 lim π‘₯β†’0+ 𝒙 𝒙 = 0 lim π‘₯β†’1βˆ’ 𝒙 𝒙 = 1 lim π‘₯β†’π‘Ž+ 𝑓 π‘₯ = f(a) and lim π‘₯β†’π‘βˆ’ 𝑓 π‘₯ = f(b) lim π‘₯β†’0+ 𝒙 𝒙 = f(0) lim π‘₯β†’1βˆ’ 𝒙 𝒙 = f(1) A function f(x) = 𝒙 𝒙 is continuous on a closed interval [0, 1].
Solution: f(x) = 𝒙 𝒙 ; [𝟎, 𝟏] lim π‘₯β†’0+ 𝒙 𝒙 = f(0) lim π‘₯β†’1βˆ’ 𝒙 𝒙 = f(1) A function f(x) = 𝒙 𝒙 is continuous on a closed interval [a, b].
Solution: f(x) = πŸ— βˆ’ π’™πŸ ; [βˆ’πŸ‘, πŸ‘] a. f(-3) = ? f(x) = πŸ— βˆ’ π’™πŸ f(-3) = πŸ— βˆ’ (βˆ’πŸ‘)𝟐 f(-3) = πŸ— βˆ’ πŸ— f(3) = ? f(x) = πŸ— βˆ’ π’™πŸ f(3) = πŸ— βˆ’ (πŸ‘)𝟐 f(3) = 𝟎 b. lim π‘₯β†’βˆ’3+ 𝑓 π‘₯ and lim π‘₯β†’3βˆ’ 𝑓 π‘₯ lim π‘₯β†’βˆ’3+ πŸ— βˆ’ π’™πŸ = πŸ— βˆ’ (βˆ’πŸ‘)𝟐 lim π‘₯β†’3βˆ’ πŸ— βˆ’ π’™πŸ = f(-3) = 𝟎 f(3) = πŸ— βˆ’ πŸ— πŸ— βˆ’ (πŸ‘)𝟐 lim π‘₯β†’βˆ’3+ πŸ— βˆ’ π’™πŸ = 0 lim π‘₯β†’3βˆ’ πŸ— βˆ’ π’™πŸ = 0
Solution: f(x) = πŸ— βˆ’ π’™πŸ; [βˆ’πŸ‘, πŸ‘] f(-3) = 𝟎 f(3) = 𝟎 lim π‘₯β†’βˆ’3+ πŸ— βˆ’ π’™πŸ = 0 lim π‘₯β†’3βˆ’ πŸ— βˆ’ π’™πŸ =0 A function f(x) = πŸ— βˆ’ π’™πŸ is continuous on a closed interval [-3,3]. lim π‘₯β†’3βˆ’ πŸ— βˆ’ π’™πŸ =f(3) lim π‘₯β†’βˆ’3+ πŸ— βˆ’ π’™πŸ = f(-3)
Solution: f(x) = πŸ— βˆ’ π’™πŸ; [βˆ’πŸ‘, πŸ‘] A function f(x) = πŸ— βˆ’ π’™πŸ is continuous on a closed interval [-3,3]. lim π‘₯β†’3βˆ’ πŸ— βˆ’ π’™πŸ = f(3) lim π‘₯β†’βˆ’3+ πŸ— βˆ’ π’™πŸ = f(-3)
Solution: f(x) = 𝟏 π’™πŸβˆ’πŸ ; [𝟎, 𝟏] a. f(0) = ? f(x) = 𝟏 π’™πŸβˆ’πŸ f(0) = 𝟏 (𝟎)πŸβˆ’πŸ f(0) = 𝟏 (𝟎)πŸβˆ’πŸ f(1) = ? f(x) = 𝟏 π’™πŸβˆ’πŸ f(1) = 𝟏 (𝟏)πŸβˆ’πŸ f(1) = undefined f(0) = βˆ’πŸ f(1) = 𝟏 𝟎 Since f(x) is not defined at x = -1 or 1; thu s, it is continuous on the open interval (0, 1).
Solution: f(x) = 𝟏 π’™πŸβˆ’πŸ ; [𝟎, 𝟏] = 𝐷𝑁𝐸 lim π‘₯β†’1βˆ’ 𝟏 π’™πŸ βˆ’ 𝟏 π‘‡β„Žπ‘’π‘ , f(x) = 𝟏 π’™πŸ βˆ’ 𝟏 π’Šπ’” 𝒏𝒐𝒕 π’„π’π’π’•π’Šπ’π’–π’π’–π’” 𝒂𝒕 [𝟎, 𝟏]

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  • 2. CONTINUITY OF A FUNCTION c c c c Graph 1 Graph 2 Graph 3 Graph 4
  • 3. CONTINUITY OF A FUNCTION c Graph 1 Graph 2 c οƒΌ The function is not defined at c οƒΌ The limit of f as x β†’ 𝒄 exist οƒΌ Both the value of the function at c and the limit as x β†’ 𝒄 exist but not equal
  • 4. CONTINUITY OF A FUNCTION c c Graph 3 Graph 4 οƒΌ The limit of f as x approaches c does not exist CONTINUOUS FUNCTION οƒΌ The value of the function is defined, that is f(c) οƒΌ The limit of f exist, that is f(c) οƒΌ The value of f and limit of f are equal
  • 5. CONTINUITY AT A POINT Definition A function f is continuous at c if and only if lim π‘₯→𝑐 𝑓 π‘₯ = 𝑓(𝑐) This implies that the three conditions must be satisfied: (1) f(c) exist; (2) lim π‘₯→𝑐 𝑓 π‘₯ exist; an d (3) lim π‘₯→𝑐 𝑓 π‘₯ = f(c)
  • 6. CONTINUITY AT A POINT Types of Discontinuity Type 1 Removable Discontinuit y 𝐼𝑓 𝑓 π‘ π‘Žπ‘‘π‘–π‘ π‘“π‘–π‘’π‘  π‘π‘œπ‘›π‘‘π‘–π‘‘π‘–π‘œπ‘› 2 , 𝑏𝑒𝑑 π‘“π‘Žπ‘–π‘™π‘  π‘‘π‘œ π‘π‘œπ‘›π‘‘π‘–π‘‘π‘–π‘œπ‘›π‘  1 π‘œπ‘Ÿ 3 , π‘‘β„Žπ‘’ π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘–π‘‘π‘¦ 𝑖𝑠 π‘π‘Žπ‘™π‘™π‘’π‘‘ π‘Ÿπ‘’π‘šπ‘œπ‘£π‘Žπ‘π‘™π‘’. π‘‡β„Žπ‘–π‘  π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘–π‘‘π‘¦ π‘œπ‘π‘π‘’π‘Ÿπ‘  π‘€β„Žπ‘’π‘› π‘‘β„Žπ‘’π‘Ÿπ‘’ 𝑖𝑠 π‘Ž β„Žπ‘œπ‘™π‘’ 𝑖𝑛 π‘‘β„Žπ‘’ π‘”π‘Ÿπ‘Žπ‘β„Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘›
  • 7. CONTINUITY AT A POINT Types of Discontinuity Type 2 Jump or Essential Discontinui ty 𝐼𝑓 𝑓 𝑖𝑠 π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘  π‘Žπ‘‘ 𝑐 π‘Žπ‘›π‘‘ π‘π‘œπ‘›π‘‘π‘–π‘‘π‘–π‘œπ‘› 2 𝑖𝑠 π‘›π‘œπ‘‘ π‘ π‘Žπ‘‘π‘–π‘ π‘“π‘–π‘’π‘‘, π‘‘β„Žπ‘’ π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘–π‘‘π‘¦ 𝑖𝑠 π‘π‘Žπ‘™π‘™π‘’π‘‘ π‘’π‘ π‘ π‘’π‘›π‘‘π‘–π‘Žπ‘™. π‘‡β„Žπ‘–π‘  π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘–π‘‘π‘¦ π‘œπ‘π‘π‘’π‘Ÿπ‘  π‘€β„Žπ‘’π‘› π‘‘β„Žπ‘’π‘”π‘Ÿπ‘Žπ‘β„Ž π‘ π‘‘π‘œπ‘π‘  π‘Žπ‘‘ π‘œπ‘›π‘’ π‘π‘œπ‘–π‘›π‘‘ π‘Žπ‘›π‘‘ π‘ π‘’π‘’π‘šπ‘‘ π‘‘π‘œ π‘—π‘’π‘šπ‘ π‘Žπ‘‘ π‘Žπ‘› π‘œπ‘‘β„Žπ‘’π‘Ÿ π‘π‘œπ‘–π‘›π‘‘. π‘‡β„Žπ‘’ 𝑙𝑒𝑓𝑑 β„Žπ‘Žπ‘›π‘‘ π‘Žπ‘›π‘‘ π‘‘β„Žπ‘’ π‘Ÿπ‘–π‘”β„Žπ‘‘ β„Žπ‘Žπ‘›π‘‘ π‘™π‘–π‘šπ‘–π‘‘π‘  𝑒π‘₯𝑖𝑠𝑑 𝑏𝑒𝑑 π‘Žπ‘Ÿπ‘’ π‘›π‘œπ‘‘ π‘’π‘žπ‘’π‘Žπ‘™.
  • 8. CONTINUITY AT A POINT Types of Discontinuity Type 3 Asymptotic or Infinite Discontinuity 𝐼𝑓 𝑓 𝑖𝑠 π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘  π‘Žπ‘‘ 𝑐 π‘Žπ‘›π‘‘ π‘π‘œπ‘›π‘‘π‘–π‘‘π‘–π‘œπ‘› 2 𝑖𝑠 π‘›π‘œπ‘‘ π‘ π‘Žπ‘‘π‘–π‘ π‘“π‘–π‘’π‘‘ π‘Žπ‘›π‘‘ π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ π‘™π‘–π‘šπ‘–π‘‘ π‘œπ‘“ π‘‘β„Žπ‘’ π‘‘π‘€π‘œ π‘™π‘–π‘šπ‘–π‘‘π‘  𝑖𝑠 𝑖𝑛𝑓𝑖𝑛𝑖𝑑𝑒 π‘‘β„Žπ‘’ π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘–π‘‘π‘¦ 𝑖𝑠 π‘π‘Žπ‘™π‘™π‘’π‘‘ 𝑖𝑛𝑓𝑖𝑛𝑖𝑑𝑒.
  • 9. EXAMPLE PRACTICE YOUR SKILLS! Given the graph of f(x),shown bel ow, determine if f(x) is continu ous at (1) x = -2 (2) x = 0 (3) x = 3
  • 10. EXAMPLE PRACTICE YOUR SKILLS! Given the graph of f(x),shown belo w, determine if f(x) is continuous at x = -2. (1) f(-2) =2 (2) lim π‘₯β†’βˆ’2 𝑓 π‘₯ does not exi st (3) lim π‘₯β†’βˆ’2 𝑓 π‘₯ β‰  f(-2) f(x) is not continuous at x = -2
  • 11. EXAMPLE PRACTICE YOUR SKILLS! Given the graph of f(x),shown belo w, determine if f(x) is continuous at x = 0. (1) f(0) =1 (2) lim π‘₯β†’0 𝑓 π‘₯ = 1, limit exis t (3) lim π‘₯β†’0 𝑓 π‘₯ = f(0) f(x) is continuous at x = 0
  • 12. EXAMPLE PRACTICE YOUR SKILLS! Given the graph of f(x),shown belo w, determine if f(x) is continuous at x = 3. (1) f(3) =-1 (2) lim π‘₯β†’3 𝑓 π‘₯ does not ex ist (3) lim π‘₯β†’3 𝑓 π‘₯ β‰  f(3) f(x) is not continuous at x = 3
  • 13. Determine whether or not the following ar e continuous functions. a. f(x) = π’™πŸ βˆ’π’™ βˆ’πŸπŸ 𝒙 βˆ’πŸ’ at x = 4 b. 𝒇 𝒙 = 𝒙 + 𝟏, π’Šπ’‡ 𝒙 < πŸ’ 𝒙 βˆ’ πŸ’ 𝟐 + πŸ‘ π’Šπ’‡ 𝒙 β‰₯ πŸ’ at x = 1 c. 𝒇 𝒙 = 𝟏 𝒙 at x = 0 EXAMPLE
  • 14. Solution f(x) = π’™πŸ βˆ’π’™βˆ’πŸπŸ 𝒙 βˆ’πŸ’ x = 4 f(4) = (πŸ’)𝟐 βˆ’(πŸ’)+𝟏𝟐 πŸ’ βˆ’πŸ’ f(4) = π’–π’π’…π’†π’‡π’Šπ’π’†π’… Condition 1
  • 15. Solution f(x) = π’™πŸ βˆ’π’™βˆ’πŸπŸ 𝒙 βˆ’πŸ’ x = 4 lim π‘₯β†’0 π’™πŸ βˆ’π’™ βˆ’ 𝟏𝟐 𝒙 βˆ’ πŸ’ Condition 2 = lim π‘₯β†’0 𝒙 βˆ’πŸ’ (𝒙+πŸ‘) 𝒙 βˆ’πŸ’ = lim π‘₯β†’0 (𝒙 + πŸ‘) = 7
  • 16. Solution f(x) = π’™πŸ βˆ’π’™βˆ’πŸπŸ 𝒙 βˆ’πŸ’ x = 4 Condition 3 lim π‘₯β†’4 𝑓 π‘₯ β‰  f(4)
  • 17. Solution f(x) = π’™πŸ βˆ’π’™βˆ’πŸπŸ 𝒙 βˆ’πŸ’ x = 4 𝐼𝑓 𝑓 π‘ π‘Žπ‘‘π‘–π‘ π‘“π‘–π‘’π‘  π‘π‘œπ‘›π‘‘π‘–π‘‘π‘–π‘œπ‘› 2 , 𝑏𝑒𝑑 π‘“π‘Žπ‘–π‘™π‘  π‘‘π‘œ π‘π‘œπ‘›π‘‘π‘–π‘‘π‘–π‘œπ‘›π‘  1 π‘œπ‘Ÿ 3 , π‘‘β„Žπ‘’ π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘–π‘‘π‘¦ 𝑖𝑠 π‘π‘Žπ‘™π‘™π‘’π‘‘ π‘Ÿπ‘’π‘šπ‘œπ‘£π‘Žπ‘π‘™π‘’.
  • 18. Solution f(x) = π’™πŸ βˆ’π’™βˆ’πŸπŸ 𝒙 βˆ’πŸ’ x = 4 𝐻𝑒𝑛𝑐𝑒, 𝑓 π‘₯ 𝑖𝑠 π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘œπ‘’π‘  π‘Žπ‘‘ π‘₯ = 4
  • 19. Solution 𝒇 𝒙 = πŸ•, 𝒙 = πŸ’ π’™πŸ βˆ’π’™βˆ’πŸπŸ 𝒙 βˆ’πŸ’ , 𝒙 β‰  πŸ’ 𝑓 π‘₯ 𝑖𝑠 π‘π‘œπ‘›π‘‘π‘–π‘›π‘œπ‘’π‘  π‘Žπ‘‘ π‘₯ = 4 Redefining the function to make it a coπ‘›π‘‘π‘–π‘›π‘œπ‘’π‘  π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘›
  • 20. Solution 𝒇 𝒙 = 𝒙 + 𝟏, π’Šπ’‡ 𝒙 < πŸ’ 𝒙 βˆ’ πŸ’ 𝟐 + πŸ‘ π’Šπ’‡ 𝒙 β‰₯ πŸ’ x = 4 f(4) = πŸ’ βˆ’ πŸ’ 𝟐 + πŸ‘ f(4) = πŸ‘ Condition 1
  • 21. Solution lim π‘₯β†’4 𝒇(𝒙) Condition 2 = 𝐷𝑁𝐸 𝒇 𝒙 = 𝒙 + 𝟏, π’Šπ’‡ 𝒙 < πŸ’ 𝒙 βˆ’ πŸ’ 𝟐 + πŸ‘ π’Šπ’‡ 𝒙 β‰₯ πŸ’ x = 4 𝐼𝑓 𝑓 𝑖𝑠 π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘  π‘Žπ‘‘ 𝑐 π‘Žπ‘›π‘‘ π‘π‘œπ‘›π‘‘π‘–π‘‘π‘–π‘œπ‘› 2 𝑖𝑠 π‘›π‘œπ‘‘ π‘ π‘Žπ‘‘π‘–π‘ π‘“π‘–π‘’π‘‘, π‘‘β„Žπ‘’ π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘–π‘‘π‘¦ 𝑖𝑠 π‘π‘Žπ‘™π‘™π‘’π‘‘ π‘’π‘ π‘ π‘’π‘›π‘‘π‘–π‘Žπ‘™. lim π‘₯β†’4βˆ’ 𝒇 𝒙 = 5 lim π‘₯β†’4+ 𝒇 𝒙 = 3
  • 22. Solution Condition 3 𝒇 𝒙 = 𝒙 + 𝟏, π’Šπ’‡ 𝒙 < πŸ’ 𝒙 βˆ’ πŸ’ 𝟐 + πŸ‘ π’Šπ’‡ 𝒙 β‰₯ πŸ’ x = 4 lim π‘₯β†’4 𝑓 π‘₯ β‰  f(4) 𝐻𝑒𝑛𝑐𝑒, 𝑓 π‘₯ 𝑖𝑠 π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘œπ‘’π‘  π‘Žπ‘‘ π‘₯ = 4
  • 23. Solution 𝒇 𝒙 = 𝟏 𝒙 𝒙 = 𝟎 f(0) = 𝟏 𝒙 f(0) = π’–π’π’…π’†π’‡π’Šπ’π’†π’… Condition 1
  • 24. Solution 𝒇 𝒙 = 𝟏 𝒙 𝒙 = 𝟎 Condition 2 lim π‘₯β†’0+ 𝒇 𝒙 = +∞ lim π‘₯β†’0βˆ’ 𝒇 𝒙 = βˆ’βˆž
  • 25. Solution 𝒇 𝒙 = 𝟏 𝒙 𝒙 = 𝟎 𝐻𝑒𝑛𝑐𝑒, 𝑓 π‘₯ 𝑖𝑠 π‘‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘œπ‘’π‘  π‘Žπ‘‘ π‘₯ = 0 and f(x) is said to have an infinite discontinuity at x = 0.
  • 26. Exercises 1. f(x) = π’™πŸ βˆ’πŸ— 𝒙 βˆ’πŸ‘ at x = 2 2. 𝒇 𝒙 = βˆ’πŸπ’™ + πŸ’, π’Šπ’‡ 𝒙 β‰₯ πŸ‘ 𝒙 βˆ’ 𝟏 π’Šπ’‡ 𝒙 < πŸ‘ at x = 3 4. 𝒇 𝒙 = π’™πŸβˆ’πŸ 𝒙 βˆ’πŸ at x = 1 3. 𝒇 𝒙 = πŸπ’™ βˆ’πŸ‘ 𝒙+ πŸ‘ at x = -3 π·π‘’π‘‘π‘’π‘Ÿπ‘šπ‘–π‘›π‘’ 𝑖𝑓 π‘‘β„Žπ‘’ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘› 𝑖𝑠 π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘  π‘Žπ‘‘ 𝑔𝑖𝑣𝑒𝑛 π‘›π‘’π‘šπ‘π‘’π‘Ÿ 𝑐.
  • 27. CONTINUITY ON AN INTERVAL Definition A function f is continuous on an open interval (a, b) if f is continuous at each number in the interv al A function f is continuous on a closed inte rval [a, b] if (𝐛) lim π‘₯β†’π‘Ž+ 𝑓 π‘₯ = f(a) and lim π‘₯β†’π‘βˆ’ 𝑓 π‘₯ = f(b) (a) f is continuous on the open interval (a, b)
  • 28. Determine if the function is continuous on the indicated closed interval. a. f(x) = 𝒙 𝒙 ; [𝟎, 𝟏] b. 𝒇 𝒙 = πŸ— βˆ’ π’™πŸ ; [-3, 3] c. 𝒇 𝒙 = 𝟏 π’™πŸ βˆ’πŸ ; [0,1] EXAMPLE
  • 29. Solution: f(x) = 𝒙 𝒙 ; [𝟎, 𝟏] a. f(a) = ? f(x) = 𝒙 𝒙 f(0) = 𝟎 𝟎 f(0) = 𝟎 f(b) = ? f(x) = 𝒙 𝒙 f(1) = 𝟏 𝟏 f(1) = 𝟏 b. lim π‘₯β†’0+ 𝑓 π‘₯ and lim π‘₯β†’1βˆ’ 𝑓 π‘₯ lim π‘₯β†’0+ 𝒙 𝒙 = 𝟎 𝟎 lim π‘₯β†’0+ 𝒙 𝒙 = 0 lim π‘₯β†’1βˆ’ 𝒙 𝒙 = 𝟏 𝟏 lim π‘₯β†’1βˆ’ 𝒙 𝒙 = 1
  • 30. Solution: f(x) = 𝒙 𝒙 ; [𝟎, 𝟏] f(0) = 𝟎 f(1) = 𝟏 lim π‘₯β†’0+ 𝒙 𝒙 = 0 lim π‘₯β†’1βˆ’ 𝒙 𝒙 = 1 lim π‘₯β†’π‘Ž+ 𝑓 π‘₯ = f(a) and lim π‘₯β†’π‘βˆ’ 𝑓 π‘₯ = f(b) lim π‘₯β†’0+ 𝒙 𝒙 = f(0) lim π‘₯β†’1βˆ’ 𝒙 𝒙 = f(1) A function f(x) = 𝒙 𝒙 is continuous on a closed interval [0, 1].
  • 31. Solution: f(x) = 𝒙 𝒙 ; [𝟎, 𝟏] lim π‘₯β†’0+ 𝒙 𝒙 = f(0) lim π‘₯β†’1βˆ’ 𝒙 𝒙 = f(1) A function f(x) = 𝒙 𝒙 is continuous on a closed interval [a, b].
  • 32. Solution: f(x) = πŸ— βˆ’ π’™πŸ ; [βˆ’πŸ‘, πŸ‘] a. f(-3) = ? f(x) = πŸ— βˆ’ π’™πŸ f(-3) = πŸ— βˆ’ (βˆ’πŸ‘)𝟐 f(-3) = πŸ— βˆ’ πŸ— f(3) = ? f(x) = πŸ— βˆ’ π’™πŸ f(3) = πŸ— βˆ’ (πŸ‘)𝟐 f(3) = 𝟎 b. lim π‘₯β†’βˆ’3+ 𝑓 π‘₯ and lim π‘₯β†’3βˆ’ 𝑓 π‘₯ lim π‘₯β†’βˆ’3+ πŸ— βˆ’ π’™πŸ = πŸ— βˆ’ (βˆ’πŸ‘)𝟐 lim π‘₯β†’3βˆ’ πŸ— βˆ’ π’™πŸ = f(-3) = 𝟎 f(3) = πŸ— βˆ’ πŸ— πŸ— βˆ’ (πŸ‘)𝟐 lim π‘₯β†’βˆ’3+ πŸ— βˆ’ π’™πŸ = 0 lim π‘₯β†’3βˆ’ πŸ— βˆ’ π’™πŸ = 0
  • 33. Solution: f(x) = πŸ— βˆ’ π’™πŸ; [βˆ’πŸ‘, πŸ‘] f(-3) = 𝟎 f(3) = 𝟎 lim π‘₯β†’βˆ’3+ πŸ— βˆ’ π’™πŸ = 0 lim π‘₯β†’3βˆ’ πŸ— βˆ’ π’™πŸ =0 A function f(x) = πŸ— βˆ’ π’™πŸ is continuous on a closed interval [-3,3]. lim π‘₯β†’3βˆ’ πŸ— βˆ’ π’™πŸ =f(3) lim π‘₯β†’βˆ’3+ πŸ— βˆ’ π’™πŸ = f(-3)
  • 34. Solution: f(x) = πŸ— βˆ’ π’™πŸ; [βˆ’πŸ‘, πŸ‘] A function f(x) = πŸ— βˆ’ π’™πŸ is continuous on a closed interval [-3,3]. lim π‘₯β†’3βˆ’ πŸ— βˆ’ π’™πŸ = f(3) lim π‘₯β†’βˆ’3+ πŸ— βˆ’ π’™πŸ = f(-3)
  • 35. Solution: f(x) = 𝟏 π’™πŸβˆ’πŸ ; [𝟎, 𝟏] a. f(0) = ? f(x) = 𝟏 π’™πŸβˆ’πŸ f(0) = 𝟏 (𝟎)πŸβˆ’πŸ f(0) = 𝟏 (𝟎)πŸβˆ’πŸ f(1) = ? f(x) = 𝟏 π’™πŸβˆ’πŸ f(1) = 𝟏 (𝟏)πŸβˆ’πŸ f(1) = undefined f(0) = βˆ’πŸ f(1) = 𝟏 𝟎 Since f(x) is not defined at x = -1 or 1; thu s, it is continuous on the open interval (0, 1).
  • 36. Solution: f(x) = 𝟏 π’™πŸβˆ’πŸ ; [𝟎, 𝟏] = 𝐷𝑁𝐸 lim π‘₯β†’1βˆ’ 𝟏 π’™πŸ βˆ’ 𝟏 π‘‡β„Žπ‘’π‘ , f(x) = 𝟏 π’™πŸ βˆ’ 𝟏 π’Šπ’” 𝒏𝒐𝒕 π’„π’π’π’•π’Šπ’π’–π’π’–π’” 𝒂𝒕 [𝟎, 𝟏]