![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
𝑥→𝑎+
𝑓 𝑥 = f(a) and lim
𝑥→𝑏−
𝑓 𝑥 = f(b)
(a) f is continuous on the open interval (a,
b)](https://image.slidesharecdn.com/continuityofafunction2023-2024-240324070414-42421c8f/85/dddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd-27-320.jpg)
![Determine if the function is continuous on
the
indicated closed interval.
a. f(x) = 𝒙 𝒙 ; [𝟎, 𝟏]
b. 𝒇 𝒙 = 𝟗 − 𝒙𝟐 ; [-3, 3]
c. 𝒇 𝒙 =
𝟏
𝒙𝟐 −𝟏
; [0,1]
EXAMPLE](https://image.slidesharecdn.com/continuityofafunction2023-2024-240324070414-42421c8f/85/dddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd-28-320.jpg)
![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
𝑥→0+
𝑓 𝑥 and lim
𝑥→1−
𝑓 𝑥
lim
𝑥→0+
𝒙 𝒙 = 𝟎 𝟎
lim
𝑥→0+
𝒙 𝒙 = 0
lim
𝑥→1−
𝒙 𝒙 = 𝟏 𝟏
lim
𝑥→1−
𝒙 𝒙 = 1](https://image.slidesharecdn.com/continuityofafunction2023-2024-240324070414-42421c8f/85/dddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd-29-320.jpg)
![Solution: f(x) = 𝒙 𝒙 ; [𝟎, 𝟏]
f(0) = 𝟎 f(1) = 𝟏
lim
𝑥→0+
𝒙 𝒙 = 0 lim
𝑥→1−
𝒙 𝒙 = 1
lim
𝑥→𝑎+
𝑓 𝑥 = f(a) and lim
𝑥→𝑏−
𝑓 𝑥 = f(b)
lim
𝑥→0+
𝒙 𝒙 = f(0) lim
𝑥→1−
𝒙 𝒙 = f(1)
A function f(x) = 𝒙 𝒙 is continuous on a
closed interval [0, 1].](https://image.slidesharecdn.com/continuityofafunction2023-2024-240324070414-42421c8f/85/dddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd-30-320.jpg)
![Solution: f(x) = 𝒙 𝒙 ; [𝟎, 𝟏]
lim
𝑥→0+
𝒙 𝒙 = f(0) lim
𝑥→1−
𝒙 𝒙 = f(1)
A function f(x) = 𝒙 𝒙 is continuous on a
closed interval [a, b].](https://image.slidesharecdn.com/continuityofafunction2023-2024-240324070414-42421c8f/85/dddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd-31-320.jpg)
![Solution: f(x) = 𝟗 − 𝒙𝟐 ; [−𝟑, 𝟑]
a. f(-3) = ?
f(x) = 𝟗 − 𝒙𝟐
f(-3) = 𝟗 − (−𝟑)𝟐
f(-3) = 𝟗 − 𝟗
f(3) = ?
f(x) = 𝟗 − 𝒙𝟐
f(3) = 𝟗 − (𝟑)𝟐
f(3) = 𝟎
b. lim
𝑥→−3+
𝑓 𝑥 and lim
𝑥→3−
𝑓 𝑥
lim
𝑥→−3+
𝟗 − 𝒙𝟐 = 𝟗 − (−𝟑)𝟐 lim
𝑥→3−
𝟗 − 𝒙𝟐 =
f(-3) = 𝟎
f(3) = 𝟗 − 𝟗
𝟗 − (𝟑)𝟐
lim
𝑥→−3+
𝟗 − 𝒙𝟐 = 0 lim
𝑥→3−
𝟗 − 𝒙𝟐 = 0](https://image.slidesharecdn.com/continuityofafunction2023-2024-240324070414-42421c8f/85/dddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd-32-320.jpg)
![Solution: f(x) = 𝟗 − 𝒙𝟐; [−𝟑, 𝟑]
f(-3) = 𝟎 f(3) = 𝟎
lim
𝑥→−3+
𝟗 − 𝒙𝟐 = 0 lim
𝑥→3−
𝟗 − 𝒙𝟐 =0
A function f(x) = 𝟗 − 𝒙𝟐 is continuous
on a closed interval [-3,3].
lim
𝑥→3−
𝟗 − 𝒙𝟐 =f(3)
lim
𝑥→−3+
𝟗 − 𝒙𝟐 = f(-3)](https://image.slidesharecdn.com/continuityofafunction2023-2024-240324070414-42421c8f/85/dddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd-33-320.jpg)
![Solution: f(x) = 𝟗 − 𝒙𝟐; [−𝟑, 𝟑]
A function f(x) = 𝟗 − 𝒙𝟐 is continuous
on a closed interval [-3,3].
lim
𝑥→3−
𝟗 − 𝒙𝟐 = f(3)
lim
𝑥→−3+
𝟗 − 𝒙𝟐 = f(-3)](https://image.slidesharecdn.com/continuityofafunction2023-2024-240324070414-42421c8f/85/dddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd-34-320.jpg)
![Solution: f(x) =
𝟏
𝒙𝟐−𝟏
; [𝟎, 𝟏]
a. f(0) = ?
f(x) =
𝟏
𝒙𝟐−𝟏
f(0) =
𝟏
(𝟎)𝟐−𝟏
f(0) =
𝟏
(𝟎)𝟐−𝟏
f(1) = ?
f(x) =
𝟏
𝒙𝟐−𝟏
f(1) =
𝟏
(𝟏)𝟐−𝟏
f(1) = undefined
f(0) = −𝟏
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).](https://image.slidesharecdn.com/continuityofafunction2023-2024-240324070414-42421c8f/85/dddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd-35-320.jpg)
![Solution:
f(x) =
𝟏
𝒙𝟐−𝟏
; [𝟎, 𝟏]
= 𝐷𝑁𝐸
lim
𝑥→1−
𝟏
𝒙𝟐 − 𝟏
𝑇ℎ𝑢𝑠, f(x) =
𝟏
𝒙𝟐 − 𝟏
𝒊𝒔 𝒏𝒐𝒕 𝒄𝒐𝒏𝒕𝒊𝒏𝒖𝒐𝒖𝒔 𝒂𝒕 [𝟎, 𝟏]](https://image.slidesharecdn.com/continuityofafunction2023-2024-240324070414-42421c8f/85/dddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd-36-320.jpg)
dddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd
- 1. MR. JOJO L. CALAGUE MATHEMATICS TEACHER
- 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 The function is not defined at c The limit of f as x → 𝒄 exist Both the value of the function at c and the limit as x → 𝒄 exist but not equal
- 4. CONTINUITY OF A FUNCTION c c Graph 3 Graph 4 The limit of f as x approaches c does not exist CONTINUOUS FUNCTION The value of the function is defined, that is f(c) The limit of f exist, that is f(c) 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 𝑥→𝑐 𝑓 𝑥 = 𝑓(𝑐) This implies that the three conditions must be satisfied: (1) f(c) exist; (2) lim 𝑥→𝑐 𝑓 𝑥 exist; an d (3) lim 𝑥→𝑐 𝑓 𝑥 = f(c)
- 6. CONTINUITY AT A POINT Types of Discontinuity Type 1 Removable Discontinuit y 𝐼𝑓 𝑓 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑠 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 2 , 𝑏𝑢𝑡 𝑓𝑎𝑖𝑙𝑠 𝑡𝑜 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠 1 𝑜𝑟 3 , 𝑡ℎ𝑒 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑟𝑒𝑚𝑜𝑣𝑎𝑏𝑙𝑒. 𝑇ℎ𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦 𝑜𝑐𝑐𝑢𝑟𝑠 𝑤ℎ𝑒𝑛 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑎 ℎ𝑜𝑙𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑔𝑟𝑎𝑝ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
- 7. CONTINUITY AT A POINT Types of Discontinuity Type 2 Jump or Essential Discontinui ty 𝐼𝑓 𝑓 𝑖𝑠 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑐 𝑎𝑛𝑑 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 2 𝑖𝑠 𝑛𝑜𝑡 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑑, 𝑡ℎ𝑒 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑒𝑠𝑠𝑒𝑛𝑡𝑖𝑎𝑙. 𝑇ℎ𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦 𝑜𝑐𝑐𝑢𝑟𝑠 𝑤ℎ𝑒𝑛 𝑡ℎ𝑒𝑔𝑟𝑎𝑝ℎ 𝑠𝑡𝑜𝑝𝑠 𝑎𝑡 𝑜𝑛𝑒 𝑝𝑜𝑖𝑛𝑡 𝑎𝑛𝑑 𝑠𝑒𝑒𝑚𝑑 𝑡𝑜 𝑗𝑢𝑚𝑝 𝑎𝑡 𝑎𝑛 𝑜𝑡ℎ𝑒𝑟 𝑝𝑜𝑖𝑛𝑡. 𝑇ℎ𝑒 𝑙𝑒𝑓𝑡 ℎ𝑎𝑛𝑑 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 ℎ𝑎𝑛𝑑 𝑙𝑖𝑚𝑖𝑡𝑠 𝑒𝑥𝑖𝑠𝑡 𝑏𝑢𝑡 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑒𝑞𝑢𝑎𝑙.
- 8. CONTINUITY AT A POINT Types of Discontinuity Type 3 Asymptotic or Infinite Discontinuity 𝐼𝑓 𝑓 𝑖𝑠 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑐 𝑎𝑛𝑑 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 2 𝑖𝑠 𝑛𝑜𝑡 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑑 𝑎𝑛𝑑 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑙𝑖𝑚𝑖𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑙𝑖𝑚𝑖𝑡𝑠 𝑖𝑠 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑒 𝑡ℎ𝑒 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑒.
- 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 𝑥→−2 𝑓 𝑥 does not exi st (3) lim 𝑥→−2 𝑓 𝑥 ≠ 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 𝑥→0 𝑓 𝑥 = 1, limit exis t (3) lim 𝑥→0 𝑓 𝑥 = 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 𝑥→3 𝑓 𝑥 does not ex ist (3) lim 𝑥→3 𝑓 𝑥 ≠ f(3) f(x) is not continuous at x = 3
- 13. Determine whether or not the following ar e continuous functions. a. f(x) = 𝒙𝟐 −𝒙 −𝟏𝟐 𝒙 −𝟒 at x = 4 b. 𝒇 𝒙 = 𝒙 + 𝟏, 𝒊𝒇 𝒙 < 𝟒 𝒙 − 𝟒 𝟐 + 𝟑 𝒊𝒇 𝒙 ≥ 𝟒 at x = 1 c. 𝒇 𝒙 = 𝟏 𝒙 at x = 0 EXAMPLE
- 14. Solution f(x) = 𝒙𝟐 −𝒙−𝟏𝟐 𝒙 −𝟒 x = 4 f(4) = (𝟒)𝟐 −(𝟒)+𝟏𝟐 𝟒 −𝟒 f(4) = 𝒖𝒏𝒅𝒆𝒇𝒊𝒏𝒆𝒅 Condition 1
- 15. Solution f(x) = 𝒙𝟐 −𝒙−𝟏𝟐 𝒙 −𝟒 x = 4 lim 𝑥→0 𝒙𝟐 −𝒙 − 𝟏𝟐 𝒙 − 𝟒 Condition 2 = lim 𝑥→0 𝒙 −𝟒 (𝒙+𝟑) 𝒙 −𝟒 = lim 𝑥→0 (𝒙 + 𝟑) = 7
- 16. Solution f(x) = 𝒙𝟐 −𝒙−𝟏𝟐 𝒙 −𝟒 x = 4 Condition 3 lim 𝑥→4 𝑓 𝑥 ≠ f(4)
- 17. Solution f(x) = 𝒙𝟐 −𝒙−𝟏𝟐 𝒙 −𝟒 x = 4 𝐼𝑓 𝑓 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑠 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 2 , 𝑏𝑢𝑡 𝑓𝑎𝑖𝑙𝑠 𝑡𝑜 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠 1 𝑜𝑟 3 , 𝑡ℎ𝑒 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑟𝑒𝑚𝑜𝑣𝑎𝑏𝑙𝑒.
- 18. Solution f(x) = 𝒙𝟐 −𝒙−𝟏𝟐 𝒙 −𝟒 x = 4 𝐻𝑒𝑛𝑐𝑒, 𝑓 𝑥 𝑖𝑠 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑜𝑢𝑠 𝑎𝑡 𝑥 = 4
- 19. Solution 𝒇 𝒙 = 𝟕, 𝒙 = 𝟒 𝒙𝟐 −𝒙−𝟏𝟐 𝒙 −𝟒 , 𝒙 ≠ 𝟒 𝑓 𝑥 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑜𝑢𝑠 𝑎𝑡 𝑥 = 4 Redefining the function to make it a co𝑛𝑡𝑖𝑛𝑜𝑢𝑠 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
- 20. Solution 𝒇 𝒙 = 𝒙 + 𝟏, 𝒊𝒇 𝒙 < 𝟒 𝒙 − 𝟒 𝟐 + 𝟑 𝒊𝒇 𝒙 ≥ 𝟒 x = 4 f(4) = 𝟒 − 𝟒 𝟐 + 𝟑 f(4) = 𝟑 Condition 1
- 21. Solution lim 𝑥→4 𝒇(𝒙) Condition 2 = 𝐷𝑁𝐸 𝒇 𝒙 = 𝒙 + 𝟏, 𝒊𝒇 𝒙 < 𝟒 𝒙 − 𝟒 𝟐 + 𝟑 𝒊𝒇 𝒙 ≥ 𝟒 x = 4 𝐼𝑓 𝑓 𝑖𝑠 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑐 𝑎𝑛𝑑 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 2 𝑖𝑠 𝑛𝑜𝑡 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑑, 𝑡ℎ𝑒 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑒𝑠𝑠𝑒𝑛𝑡𝑖𝑎𝑙. lim 𝑥→4− 𝒇 𝒙 = 5 lim 𝑥→4+ 𝒇 𝒙 = 3
- 22. Solution Condition 3 𝒇 𝒙 = 𝒙 + 𝟏, 𝒊𝒇 𝒙 < 𝟒 𝒙 − 𝟒 𝟐 + 𝟑 𝒊𝒇 𝒙 ≥ 𝟒 x = 4 lim 𝑥→4 𝑓 𝑥 ≠ f(4) 𝐻𝑒𝑛𝑐𝑒, 𝑓 𝑥 𝑖𝑠 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑜𝑢𝑠 𝑎𝑡 𝑥 = 4
- 23. Solution 𝒇 𝒙 = 𝟏 𝒙 𝒙 = 𝟎 f(0) = 𝟏 𝒙 f(0) = 𝒖𝒏𝒅𝒆𝒇𝒊𝒏𝒆𝒅 Condition 1
- 24. Solution 𝒇 𝒙 = 𝟏 𝒙 𝒙 = 𝟎 Condition 2 lim 𝑥→0+ 𝒇 𝒙 = +∞ lim 𝑥→0− 𝒇 𝒙 = −∞
- 25. Solution 𝒇 𝒙 = 𝟏 𝒙 𝒙 = 𝟎 𝐻𝑒𝑛𝑐𝑒, 𝑓 𝑥 𝑖𝑠 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑜𝑢𝑠 𝑎𝑡 𝑥 = 0 and f(x) is said to have an infinite discontinuity at x = 0.
- 26. Exercises 1. f(x) = 𝒙𝟐 −𝟗 𝒙 −𝟑 at x = 2 2. 𝒇 𝒙 = −𝟐𝒙 + 𝟒, 𝒊𝒇 𝒙 ≥ 𝟑 𝒙 − 𝟏 𝒊𝒇 𝒙 < 𝟑 at x = 3 4. 𝒇 𝒙 = 𝒙𝟐−𝟏 𝒙 −𝟏 at x = 1 3. 𝒇 𝒙 = 𝟐𝒙 −𝟑 𝒙+ 𝟑 at x = -3 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑖𝑓 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑔𝑖𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 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 𝑥→𝑎+ 𝑓 𝑥 = f(a) and lim 𝑥→𝑏− 𝑓 𝑥 = 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 = 𝟗 − 𝒙𝟐 ; [-3, 3] c. 𝒇 𝒙 = 𝟏 𝒙𝟐 −𝟏 ; [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 𝑥→0+ 𝑓 𝑥 and lim 𝑥→1− 𝑓 𝑥 lim 𝑥→0+ 𝒙 𝒙 = 𝟎 𝟎 lim 𝑥→0+ 𝒙 𝒙 = 0 lim 𝑥→1− 𝒙 𝒙 = 𝟏 𝟏 lim 𝑥→1− 𝒙 𝒙 = 1
- 30. Solution: f(x) = 𝒙 𝒙 ; [𝟎, 𝟏] f(0) = 𝟎 f(1) = 𝟏 lim 𝑥→0+ 𝒙 𝒙 = 0 lim 𝑥→1− 𝒙 𝒙 = 1 lim 𝑥→𝑎+ 𝑓 𝑥 = f(a) and lim 𝑥→𝑏− 𝑓 𝑥 = f(b) lim 𝑥→0+ 𝒙 𝒙 = f(0) lim 𝑥→1− 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 𝑥→0+ 𝒙 𝒙 = f(0) lim 𝑥→1− 𝒙 𝒙 = f(1) A function f(x) = 𝒙 𝒙 is continuous on a closed interval [a, b].
- 32. Solution: f(x) = 𝟗 − 𝒙𝟐 ; [−𝟑, 𝟑] a. f(-3) = ? f(x) = 𝟗 − 𝒙𝟐 f(-3) = 𝟗 − (−𝟑)𝟐 f(-3) = 𝟗 − 𝟗 f(3) = ? f(x) = 𝟗 − 𝒙𝟐 f(3) = 𝟗 − (𝟑)𝟐 f(3) = 𝟎 b. lim 𝑥→−3+ 𝑓 𝑥 and lim 𝑥→3− 𝑓 𝑥 lim 𝑥→−3+ 𝟗 − 𝒙𝟐 = 𝟗 − (−𝟑)𝟐 lim 𝑥→3− 𝟗 − 𝒙𝟐 = f(-3) = 𝟎 f(3) = 𝟗 − 𝟗 𝟗 − (𝟑)𝟐 lim 𝑥→−3+ 𝟗 − 𝒙𝟐 = 0 lim 𝑥→3− 𝟗 − 𝒙𝟐 = 0
- 33. Solution: f(x) = 𝟗 − 𝒙𝟐; [−𝟑, 𝟑] f(-3) = 𝟎 f(3) = 𝟎 lim 𝑥→−3+ 𝟗 − 𝒙𝟐 = 0 lim 𝑥→3− 𝟗 − 𝒙𝟐 =0 A function f(x) = 𝟗 − 𝒙𝟐 is continuous on a closed interval [-3,3]. lim 𝑥→3− 𝟗 − 𝒙𝟐 =f(3) lim 𝑥→−3+ 𝟗 − 𝒙𝟐 = f(-3)
- 34. Solution: f(x) = 𝟗 − 𝒙𝟐; [−𝟑, 𝟑] A function f(x) = 𝟗 − 𝒙𝟐 is continuous on a closed interval [-3,3]. lim 𝑥→3− 𝟗 − 𝒙𝟐 = f(3) lim 𝑥→−3+ 𝟗 − 𝒙𝟐 = f(-3)
- 35. Solution: f(x) = 𝟏 𝒙𝟐−𝟏 ; [𝟎, 𝟏] a. f(0) = ? f(x) = 𝟏 𝒙𝟐−𝟏 f(0) = 𝟏 (𝟎)𝟐−𝟏 f(0) = 𝟏 (𝟎)𝟐−𝟏 f(1) = ? f(x) = 𝟏 𝒙𝟐−𝟏 f(1) = 𝟏 (𝟏)𝟐−𝟏 f(1) = undefined f(0) = −𝟏 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 𝒙𝟐−𝟏 ; [𝟎, 𝟏] = 𝐷𝑁𝐸 lim 𝑥→1− 𝟏 𝒙𝟐 − 𝟏 𝑇ℎ𝑢𝑠, f(x) = 𝟏 𝒙𝟐 − 𝟏 𝒊𝒔 𝒏𝒐𝒕 𝒄𝒐𝒏𝒕𝒊𝒏𝒖𝒐𝒖𝒔 𝒂𝒕 [𝟎, 𝟏]