kjdhEWOIH

1. Calculus and
Analytical
Geometry
Semester-1
Amna Tahir
08-12-2020

2. 2

3. Today’s Outline
• Revision of continuity (Ex: 1.3)
• Revision of one-sided derivatives (Ex: 2.1)
3
Absolute value function
𝑓 𝑥 = ቊ
𝑥, 𝑥 ≥ 0
−𝑥, 𝑥 < 0

4. Continuity of Absolute functions
Question-1:
Discuss the continuity of 𝑓 𝑥 = 𝑥 − 3 at 𝑥 = 3.
Solution:
• Functional value:
Given, 𝑓 𝑥 = 𝑥 − 3
At 𝑥 = 3, 𝑓 3 = 3 − 3 = 0 = 0
𝑓 3 = 0
• Existence of limit:
lim
𝑥→3
𝑓 𝑥 = lim
𝑥→3
𝑥 − 3
4

5. Continuity of Absolute functions
As our given function is absolute value function so according to the
definition of absolute function, we write 𝑓(𝑥) as;
𝑓 𝑥 = 𝑥 − 3
𝑓 𝑥 = 𝑥 − 3 = ቊ
𝑥 − 3 𝑖𝑓 𝑥 − 3 ≥ 0
− 𝑥 − 3 𝑖𝑓 𝑥 − 3 < 0
𝑓 𝑥 = 𝑥 − 3 = ቊ
𝑥 − 3 𝑖𝑓 𝑥 ≥ 3
− 𝑥 − 3 𝑖𝑓 𝑥 < 3
5

6. Continuity of Absolute functions
Left hand limit
lim
𝑥→3−
𝑓(𝑥) = lim
𝑥→3−
− 𝑥 − 3
lim
𝑥→3−
𝑓(𝑥) = lim
𝑥→3−
−𝑥 + 3
lim
𝑥→3−
𝑓 𝑥 = − lim
𝑥→3−
𝑥 + lim
𝑥→3−
3
lim
𝑥→3−
𝑓 𝑥 = −3 + 3 = 0
Right hand limit
lim
𝑥→3+
𝑓 𝑥 = lim
𝑥→3+
𝑥 − 3
lim
𝑥→3+
𝑓 𝑥 = lim
𝑥→3+
𝑥 − lim
𝑥→3+
3
lim
𝑥→3+
𝑓 𝑥 = 3 − 3 = 0
6

7. Continuity of Absolute functions
As both LHL and RHL are equal i.e.
lim
𝑥→3−
𝑓 𝑥 = 0 = lim
𝑥→3+
𝑓 𝑥
⟹ lim
𝑥→3
𝑓(𝑥) exists and also equal to zero.
⟹ lim
𝑥→3
𝑓 𝑥 = 0
• From above conditions, we have
𝑓 3 = 0 = lim
𝑥→3
𝑓 𝑥
Hence all the three conditions are satisfied so 𝑓(𝑥) is continuous at
𝑥 = 3.
7

8. Continuity of functions
Question-2:
Discuss the continuity of 𝑥 − 𝑥 at 𝑥 = 1.
Solution:
• Functional value:
Given, 𝑓 𝑥 = 𝑥 − 𝑥
At 𝑥 = 1, 𝑓 1 = 1 − 1 = 1 − 1 = 0
𝑓 1 = 0
• Existence of limit:
lim
𝑥→1
𝑓 𝑥 = lim
𝑥→1
( 𝑥 − 𝑥 )
8

9. Continuity of functions
As our given function contains absolute value function so according to
the definition of absolute function, we write 𝑓(𝑥) as;
𝑓 𝑥 = x − 𝑥 = ቊ
𝑥 − 𝑥 𝑖𝑓 𝑥 ≥ 0
𝑥 − −𝑥 𝑖𝑓 𝑥 < 0
= ቊ
0 𝑖𝑓 𝑥 ≥ 0
2𝑥 𝑖𝑓 𝑥 < 0
But we have to discuss the continuity of 𝑓(𝑥) at 𝑥 = 1, so we re-arrange
the above function for 𝑥 = 1.
𝑓 𝑥 = x − 𝑥 = ൞
0 𝑖𝑓 𝑥 ≥ 1
0 𝑖𝑓 0 < 𝑥 < 1
2𝑥 𝑖𝑓 𝑥 < 0
9

10. Continuity of functions
Left-hand limit lim
𝑥→1−
𝑓(𝑥) = lim
𝑥→1−
0 = 0
Right-hand limit lim
𝑥→1+
𝑓(𝑥) = lim
𝑥→1+
0 = 0
As both LHL and RHL are equal i.e.
lim
𝑥→1−
𝑓 𝑥 = 0 = lim
𝑥→1+
𝑓 𝑥
⟹ lim
𝑥→1
𝑓(𝑥) exists and also equal to zero.
⟹ lim
𝑥→1
𝑓 𝑥 = 0
10

11. Continuity of functions
• From above two conditions, we have
𝑓 1 = 0 = lim
𝑥→1
𝑓 𝑥
Hence all the three conditions are satisfied so 𝑓(𝑥) is continuous at
𝑥 = 1.
Note:
In Q-2, given function 𝑓 𝑥 = 𝑥 − 𝑥 is a difference of polynomial and
absolute functions so it must be continuous at every point.
11

12. One-Sided Derivatives
12

13. One-sided derivatives
Left-hand derivative
• 𝐿𝑓′
(x) = lim
ℎ→0−
𝑓 𝑥+ℎ −𝑓(𝑥)
ℎ
OR
• 𝐿𝑓′(a) = lim
𝑥→𝑎−
𝑓 𝑥 −𝑓(𝑎)
𝑥−𝑎
Right-hand derivative
• 𝑅𝑓′(x) = lim
ℎ→0+
𝑓 𝑥+ℎ −𝑓(𝑥)
ℎ
OR
• 𝑅𝑓′(a) = lim
𝑥→𝑎+
𝑓 𝑥 −𝑓(𝑎)
𝑥−𝑎
13

14. One-sided derivatives
Question-3:
Find 𝐿𝑓′
(2) and 𝑅𝑓′
(2) for the
function 𝑓 𝑥 = 𝑥2 − 4 .
Solution:
∵ 𝐿𝑓′(a) = lim
𝑥→𝑎−
𝑓 𝑥 −𝑓(𝑎)
𝑥−𝑎
𝐿𝑓′(2) = lim
𝑥→2−
𝑓 𝑥 −𝑓(2)
𝑥−2
𝐿𝑓′(2) = lim
𝑥→2−
𝑥2−4 − (2)2−4
𝑥−2
𝐿𝑓′(2) = lim
𝑥→2−
−(𝑥2−4)− 4−4
𝑥−2
𝐿𝑓′(2) = lim
𝑥→2−
−(𝑥2−22)− 0
𝑥−2
𝐿𝑓′
(2) = lim
𝑥→2−
−(𝑥−2)(𝑥+2)
𝑥−2
𝐿𝑓′
(2) = lim
𝑥→2−
− 𝑥 + 2
𝐿𝑓′(2) = − 2 + 2 = −4
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15. One-sided derivatives
Question-3: (𝑓 𝑥 = 𝑥2 − 4 )
• When 𝒙 → 𝟐− , it means 𝑥 < 2.
⟹ 𝑥2
< 4
(Taking square on both sides)
⟹ 𝑥2 − 4 < 0
⟹ 𝑥2 − 4 = −(𝑥2−4)
• When 𝒙 → 𝟐+
, it means 𝑥 > 2.
⟹ 𝑥2 > 4
(Taking square on both sides)
⟹ 𝑥2 − 4 > 0
⟹ 𝑥2
− 4 = 𝑥2
− 4
15

16. One-sided derivatives
• Alternate approach
𝑓 𝑥 = 𝑥2 − 4 = ൝
𝑥2
−4 𝑖𝑓𝑥2
− 4 ≥ 0
−(𝑥2 − 4) 𝑖𝑓𝑥2 − 4 < 0
𝑓 𝑥 = 𝑥2 − 4 = ൝
𝑥2
−4 𝑖𝑓𝑥2
≥ 4
−(𝑥2
− 4) 𝑖𝑓𝑥2
< 4
𝑓 𝑥 = 𝑥2
− 4 = ൝
𝑥2−4 𝑖𝑓𝑥 ≥ 2
−(𝑥2 − 4) 𝑖𝑓𝑥 < 2
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17. Differentiability & Continuity
Question-4:
Let 𝑓 𝑥 = ቐ
𝑥𝑡𝑎𝑛ℎ
1
𝑥
𝑖𝑓 𝑥 ≠ 0
0 𝑖𝑓 𝑥 = 0
Discuss the continuity and differentiability of 𝑓 at 𝑥 = 0.
Solution:
Continuity already discussed in previous lecture.
• Differentiability of 𝒇
𝑓′
(0) = lim
𝑥→0
𝑓 𝑥 −𝑓(0)
𝑥−0
= lim
𝑥→0
𝑥𝑡𝑎𝑛ℎ
1
𝑥
− (0)𝑡𝑎𝑛ℎ
1
0
𝑥
17

18. Differentiability & Continuity
𝑓′(0) = lim
𝑥→0
𝑥𝑡𝑎𝑛ℎ
1
𝑥
− 0
𝑥
= lim
𝑥→0
𝑥𝑡𝑎𝑛ℎ
1
𝑥
𝑥
= lim
𝑥→0
𝑡𝑎𝑛ℎ
1
𝑥
As 𝑡𝑎𝑛ℎ𝑥 =
𝑒𝑥 − 𝑒−𝑥
𝑒𝑥 + 𝑒−𝑥 ⟹ 𝑡𝑎𝑛ℎ
1
𝑥
=
𝑒1/𝑥 − 𝑒−1/𝑥
𝑒1/𝑥 + 𝑒−1/𝑥
So, 𝑓′(0) = lim
𝑥→0
𝑡𝑎𝑛ℎ
1
𝑥
= lim
𝑥→0
𝑒1/𝑥 − 𝑒−1/𝑥
𝑒1/𝑥 + 𝑒−1/𝑥
Because of exponential function, we have to discuss two cases i.e.
𝑥 → 0−
and 𝑥 → 0+
.
18

19. Differentiability & Continuity
𝐿𝑓′
(0) = lim
𝑥→0−
𝑒1/𝑥 − 𝑒−1/𝑥
𝑒1/𝑥 + 𝑒−1/𝑥
𝐿𝑓′
(0) = lim
𝑥→0−
𝑒1/𝑥 −
1
𝑒1/𝑥
𝑒1/𝑥 +
1
𝑒1/𝑥
𝐿𝑓′(0) = lim
𝑥→0−
𝑒2/𝑥 − 1
𝑒1/𝑥
𝑒2/𝑥 + 1
𝑒1/𝑥
= lim
𝑥→0−
𝑒2/𝑥 − 1
𝑒2/𝑥 + 1
=
0 − 1
0 + 1
𝐿𝑓′
(0) = − 1
19
As 𝑥 → 0−,
1
𝑥
→ −∞
and 𝑒1/𝑥 → 0
But 𝑒−1/𝑥 → ∞
We have to remove
those terms which
became undefined.

20. Differentiability & Continuity
𝑅𝑓′(0)= lim
𝑥→0+
𝑒1/𝑥 − 𝑒−1/𝑥
𝑒1/𝑥 + 𝑒−1/𝑥
𝑅𝑓′
(0)= lim
𝑥→0+
𝑒1/𝑥 (1 − 𝑒
−
2
𝑥)
𝑒1/𝑥 (1 + 𝑒
−
2
𝑥)
𝑅𝑓′(0)= lim
𝑥→0+
1 − 𝑒
−
2
𝑥
1 + 𝑒
−
2
𝑥
=
1 − 0
1 + 0
𝑅𝑓′(0) =1
Hence 𝐿𝑓′(0) ≠ 𝑅𝑓′(0) ⟹ 𝑓′ 0 does not exist.
20
As 𝑥 → 0+,
1
𝑥
→ +∞
and 𝑒1/𝑥
→ ∞
But 𝑒−1/𝑥 → 0
We have to remove
those terms which
became undefined.

21. Techniques of Differentiation
21

22. General Theorems on Derivatives
• The derivative of a constant function (i.e. 𝑓: 𝑥 → 𝑐 defined by,
𝑓(𝑥) = 𝑐) is zero.
⟹
𝑑
𝑑𝑥
𝑐 = 0
e.g.
𝑑
𝑑𝑥
(𝜋) = 0,
𝑑
𝑑𝑥
−5 = 0
• The derivative of an identity function (i.e. 𝑓: 𝑥 → 𝑥 defined by,
𝑓(𝑥) = 𝑥) is one.
⟹
𝑑
𝑑𝑥
𝑥 = 1
e.g.
𝑑
𝑑𝑦
(𝑦) = 1,
𝑑
𝑑𝑡
𝑡 = 1
22

23. General Theorems on Derivatives
• The power rule
If 𝑛 is a any real number, then
𝑑
𝑑𝑥
𝑥𝑛 = 𝑛𝑥𝑛−1.
e.g.
𝑑
𝑑𝑥
𝑥−9
= −9𝑥−9−1
= −9 𝑥−10
• A constant factor can be moved through a derivative sign.
⟹
𝑑
𝑑𝑥
𝑐𝑓 𝑥 = c
𝑑
𝑑𝑥
[𝑓(𝑥)]
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