Mathematics Department

Dr. Mohammad Yasin

Worksheet (1) Solution

Maclaurin and Taylor Series

1. Apply the definition in order to find Maclaurin/Taylor series of each of the following functions:

1
a. 𝑓(𝑥) = cosh 𝑥 c. 𝑓(𝑥) = , 𝑎=1 e. 𝑓(𝑥) = 5 cos 𝜋𝑥
𝑥2

𝑥
b. 𝑓(𝑥) = d. 𝑓(𝑥) = ln 𝑥, 𝑎=2 f. 𝑓(𝑥) = ln √𝑥 + 1
𝑥+1

Solution:

𝐚. 𝒇(𝒙) = 𝐜𝐨𝐬𝐡 𝒙

Maclaurin series of the function 𝑓(𝑥) is given by

′ (0) ′′ (0)
𝑥2 ′′′ (0)
𝑥3 (𝑛)
𝑥𝑛
𝑓(𝑥) = 𝑓(0) + 𝑓 𝑥+𝑓 +𝑓 + ⋯ + 𝑓 (0) +⋯
2! 3! 𝑛!

𝑓(𝑥) = cosh 𝑥 𝑓(0) = 1

𝑓 ′ (𝑥) = sinh 𝑥 𝑓 ′ (0) = 0

𝑓 ′′ (𝑥) = cosh 𝑥 𝑓 ′′ (0) = 1

𝑓 ′′′ (𝑥) = sinh 𝑥 𝑓 ′′′ (0) = 0

𝑥2 𝑥4
∴ cosh 𝑥 = 1 + + + ⋯ ⋯
2! 4!

Page 1 of 13
 𝒙
𝐛. 𝒇(𝒙) =
𝒙+𝟏
𝑥 1
𝑓(𝑥) = = 𝑥( )
𝑥+1 𝑥+1

Starting with the function 𝒈(𝒙) = 𝟏⁄(𝒙 + 𝟏) to get the derivatives easily

𝑔(𝑥) = 1⁄(𝑥 + 1) 𝑔(0) = 1

𝑔′(𝑥) = −1⁄(𝑥 + 1)2 𝑔′ (0) = −1

𝑔′′(𝑥) = 2⁄(𝑥 + 1)3 𝑔′′ (0) = 2

𝑔′′′(𝑥) = −6⁄(𝑥 + 1)4 𝑔′′′ (0) = −6

1
∴ = 1 − 𝑥 + 𝑥2 − 𝑥3 + ⋯
1+𝑥
Multiply by 𝒙 you’ll get:
𝑥
∴ = 𝑥 − 𝑥2 + 𝑥3 − 𝑥4 + ⋯ ⋯
𝑥+1
Another Approach:
1
Recall that = 1 − 𝑥 + 𝑥2 − 𝑥3 + ⋯ ⋯
1+𝑥

Multiply the series for 𝟏⁄(𝟏 + 𝒙) by 𝒙.

𝑥
∴ = 𝑥 − 𝑥2 + 𝑥3 − 𝑥4 + ⋯ ⋯
𝑥+1

𝟏
𝐜. 𝒇(𝒙) = , 𝒂=𝟏
𝒙𝟐

Taylor series of the function 𝑓(𝑥) at 𝑥 = 𝑎 is given by

′ (𝑎) ′′ (𝑎)
(𝑥 − 𝑎)2 (𝑛)
(𝑥 − 𝑎)𝑛
𝑓(𝑥) = 𝑓(𝑎) + 𝑓 (𝑥 − 𝑎) + 𝑓 + ⋯ + 𝑓 (𝑎) +⋯
2! 𝑛!

Page 2 of 13
 𝑓(𝑥) = 1⁄𝑥 2 𝑓(1) = 1

𝑓 ′ (𝑥) = −2⁄𝑥 3 𝑓 ′ (1) = −2

𝑓 ′′ (𝑥) = 6⁄𝑥 4 𝑓 ′′ (1) = 6

𝑓 ′′′ (𝑥) = −24⁄𝑥 5 𝑓 ′′′ (1) = −24

1 6 2
24
∴ 2
= 1 − 2(𝑥 − 1) + (𝑥 − 1) − (𝑥 − 1)3 + ⋯ ⋯
𝑥 2! 3!

𝐝. 𝒇(𝒙) = 𝐥𝐧 𝒙, 𝒂 = 𝟐

𝑓(𝑥) = ln 𝑥 𝑓(2) = ln 2

1 1
𝑓 ′ (𝑥) = 𝑓 ′ (2) =
𝑥 2
−1 −1
𝑓 ′′ (𝑥) = 𝑓 ′′ (2) =
𝑥2 4
2 1
𝑓 ′′′ (𝑥) = 𝑓 ′′′ (2) =
𝑥3 4
−1 1
1 (4) (4)
2 (𝑥 − 2)3 + ⋯ ⋯
∴ ln 𝑥 = ln 2 + ( ) (𝑥 − 2) + (𝑥 − 2) +
2 2! 3!

𝐞. 𝒇(𝒙) = 𝟓 𝐜𝐨𝐬 𝝅𝒙

𝑓(𝑥) = 5 cos(𝜋𝑥) 𝑓(0) = 5

𝑓 ′ (𝑥) = −5𝜋 sin(𝜋𝑥) 𝑓 ′ (0) = 0

𝑓 ′′ (𝑥) = −5𝜋 2 cos(𝜋𝑥) 𝑓 ′′ (0) = −5𝜋 2

𝑓 ′′′ (𝑥) = 5𝜋 3 sin(𝜋𝑥) 𝑓 ′′′ (0) = 0

(−5𝜋 2 ) 2
∴ 5 cos(𝜋𝑥) = 5 + 𝑥 + ⋯⋯
2!

Page 3 of 13
Another Approach:
It's easier to replace every (𝒙) by (𝝅𝒙) in the expansion of 𝐜𝐨𝐬 𝒙.

1 2 1 4
Recall that cos 𝑥 = 1 − 𝑥 + 𝑥 − ⋯⋯
2! 4!

1 1
cos(𝜋𝑥) = 1 − (𝜋𝑥)2 + (𝜋𝑥)4 − ⋯ ⋯
2! 4!

Multiply the series for 𝐜𝐨𝐬(𝝅𝒙) by 𝟓.

1 1
∴ 5 cos(𝜋𝑥) = 5 [1 − (𝜋𝑥)2 + (𝜋𝑥)4 − ⋯ ⋯ ]
2! 4!

𝟏
𝐟. 𝒇(𝒙) = 𝐥𝐧 √𝒙 + 𝟏 = 𝐥𝐧(𝒙 + 𝟏)
𝟐

1
𝑓(𝑥) = ln(𝑥 + 1) 𝑓(0) = 0
2
1 1
𝑓 ′ (𝑥) = 𝑓 ′ (0) =
2(𝑥 + 1) 2
−1 −1
𝑓 ′′ (𝑥) = 𝑓 ′′ (0) =
2(𝑥 + 1)2 2
1
𝑓 ′′′ (𝑥) = 𝑓 ′′′ (0) = 1
(𝑥 + 1)3

1
1 (− 2) 1
∴ ln √𝑥 + 1 = ( ) 𝑥 + 𝑥2 + 𝑥3 + ⋯ ⋯
2 2! 3!

Another Approach:
1
Recall that = 1 − 𝑥 + 𝑥2 − 𝑥3 + ⋯ ⋯
1+𝑥
Integrate the series for 𝟏⁄(𝟏 + 𝒙) with respect to 𝒙.

𝑥2 𝑥3 𝑥4
ln(𝑥 + 1) = 𝑥 − + − + ⋯ ⋯ + 𝑐
2 3 4

Page 4 of 13
Put 𝑥 = 0 in both sides, we get

ln 1 = 𝑐 ⟹ 𝑐=0

Multiply the series for 𝐥𝐧(𝒙 + 𝟏) by 𝟏⁄𝟐.

1 1 𝑥2 𝑥3 𝑥4
∴ ln(𝑥 + 1) = [𝑥 − + − + ⋯ ⋯ ]
2 2 2 3 4

2. Find a power series representation for each of the following functions:

5 𝑥2
a. 𝑓(𝑥) = c. 𝑓(𝑥) =
1 − 4𝑥 2 8 − 𝑥3

1+𝑥 1
b. 𝑓(𝑥) = d. 𝑓(𝑥) =
1−𝑥 (1 + 𝑥)2

Solution:

𝟓
𝐚. 𝒇(𝒙) =
𝟏 − 𝟒𝒙𝟐

1
Recall that = 1 + 𝑥 + 𝑥2 + 𝑥3 + ⋯ ⋯
1−𝑥

Replace every (𝒙) by (𝟒𝒙𝟐 ) in the expansion of 𝟏⁄(𝟏 − 𝒙)

1
= 1 + (4𝑥 2 ) + (4𝑥 2 )2 + (4𝑥 2 )3 + ⋯ = 1 + 4𝑥 2 + 16𝑥 4 + 64𝑥 6 + ⋯
1 − 4𝑥 2

Multiply the series for 𝟏⁄(𝟏 − 𝟒𝒙𝟐 ) by 𝟓.

5
∴ = 5 + 20𝑥 2 + 80𝑥 4 + 320𝑥 6 + ⋯
1 − 4𝑥 2

Page 5 of 13
 𝟏+𝒙
𝐛. 𝒇(𝒙) =
𝟏−𝒙
1+𝑥 1 1
𝑓(𝑥) = = +𝑥( )
1−𝑥 1−𝑥 1−𝑥

1
Recall that = 1 + 𝑥 + 𝑥2 + 𝑥3 + ⋯ ⋯
1−𝑥

Multiply the series for 𝟏⁄(𝟏 − 𝒙) by 𝒙.

1
𝑥( ) = 𝑥 + 𝑥2 + 𝑥3 + 𝑥4 …
1−𝑥

1+𝑥 1 1
= +𝑥( ) = (1 + 𝑥 + 𝑥 2 + 𝑥 3 … ) + (𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4 … )
1−𝑥 1−𝑥 1−𝑥

1+𝑥
∴ = 1 + 2𝑥 + 2𝑥 2 + 2𝑥 3 + ⋯
1−𝑥

𝒙𝟐
𝐜. 𝒇(𝒙) =
𝟖 − 𝒙𝟑

𝑥2 𝑥2 𝑥2 1
𝑓(𝑥) = = 3 = ( ) [ 𝑥3
]
8 − 𝑥 3 8 [1 − (𝑥 )] 8 1−( )
8 8

1
Recall that = 1 + 𝑥 + 𝑥2 + 𝑥3 + ⋯ ⋯
1−𝑥

𝒙𝟑
Replace every (𝒙) by ( ) in the expansion of 𝟏⁄(𝟏 − 𝒙)
𝟖

2 3
1 𝑥3 𝑥3 𝑥3 𝑥3 𝑥6 𝑥9
𝑥3
= 1 + ( ) + [( )] + [( )] + ⋯ = 1 + + + +⋯
1−( ) 8 8 8 8 64 512
8

𝑥2 𝑥2 1 𝑥2 𝑥5 𝑥8 𝑥 11
∴ =( ) [ 𝑥3
]= + + + +⋯
8 − 𝑥3 8 1−( ) 8 64 512 4096
8

Page 6 of 13
 𝟏
𝐝. 𝒇(𝒙) =
(𝟏 + 𝒙)𝟐
1
Recall that = 1 − 𝑥 + 𝑥2 − 𝑥3 + ⋯ ⋯
1+𝑥

Differentiate the series for 𝟏⁄(𝟏 + 𝒙) with respect to 𝒙.

−1
= −1 + 2𝑥 − 3𝑥 2 + ⋯
(1 + 𝑥)2

Multiply the series for −𝟏⁄(𝟏 + 𝒙𝟐 )𝟐 by −𝟏.

1 −1
∴ = − [ ] = 1 − 2𝑥 + 3𝑥 2 − ⋯
(1 + 𝑥)2 (1 + 𝑥)2

Another Approach: You can use the binomial theorem for (𝟏 + 𝒙)−𝟐 .

3. Find Maclaurin series of each of the following functions:

𝑥
sin 𝑥 d. 𝑓(𝑥) =
a. 𝑓(𝑥) = 𝑒
(1 + 4𝑥 2 )2

2
b. 𝑓(𝑥) = 𝑒 −𝑥 + cos 𝑥 e. 𝑓(𝑥) = 𝑥 2 tan−1 (𝑥 3 )

c. 𝑓(𝑥) = 𝑥 2 + sinh(𝑥 4 )

Solution:

𝐚. 𝒇(𝒙) = 𝒆𝐬𝐢𝐧 𝒙

𝑓(𝑥) = 𝑒 sin 𝑥 𝑓(0) = 1

𝑓 ′ (𝑥) = 𝑒 sin 𝑥 cos 𝑥 𝑓 ′ (0) = 1

𝑓 ′′ (𝑥) = 𝑒 sin 𝑥 (− sin 𝑥) + 𝑒 sin 𝑥 cos 2 𝑥 𝑓 ′′ (0) = 1

sin 𝑥
𝑥2
∴𝑒 =1+𝑥+ +⋯
2!

Page 7 of 13
 𝟐
𝐛. 𝒇(𝒙) = 𝒆−𝒙 + 𝐜𝐨𝐬 𝒙

1 2 1 3
Recall that 𝑒𝑥 = 1 + 𝑥 + 𝑥 + 𝑥 +⋯⋯
2! 3!

Replace every (𝒙) by (−𝒙𝟐 ) in the expansion of 𝒆𝒙

2 1 1 1 1
𝑒 −𝑥 = 1 + (−𝑥 2 ) + (−𝑥 2 )2 + (−𝑥 2 )3 + ⋯ = 1 − 𝑥 2 + 𝑥 4 − 𝑥 6 + ⋯
2! 3! 2! 3!

1 2 1 4 1 6
Recall that cos 𝑥 = 1 − 𝑥 + 𝑥 − 𝑥 +⋯
2! 4! 6!

2 1 1 1 1 1 6
∴ 𝑒 −𝑥 + cos 𝑥 = (1 − 𝑥 2 + 𝑥 4 − 𝑥 6 + ⋯ ) + (1 − 𝑥 2 + 𝑥 4 − 𝑥 + ⋯)
2 6 2 24 720

3 13 121 6
= 2 − 𝑥2 + 𝑥4 − 𝑥 +⋯
2 24 720

𝐜. 𝒇(𝒙) = 𝒙𝟐 + 𝐬𝐢𝐧𝐡(𝒙𝟒 )

1 3 1 5
Recall that sinh 𝑥 = 𝑥 + 𝑥 + 𝑥 +⋯
3! 5!

Replace every (𝒙) by (𝒙𝟒 ) in the expansion of 𝐬𝐢𝐧𝐡 𝒙

1 4 3 1 4 5
sinh(𝑥 4 ) = (𝑥 4 ) + (𝑥 ) + (𝑥 ) + ⋯
3! 5!

2 4) 2
𝑥 12 𝑥 20
4
∴ 𝑥 + sinh(𝑥 =𝑥 +𝑥 + + +⋯
3! 5!

𝒙
𝐝. 𝒇(𝒙) =
(𝟏 + 𝟒𝒙𝟐 )𝟐
1
Recall that = 1 − 𝑥 + 𝑥2 − 𝑥3 + ⋯ ⋯
1+𝑥

Page 8 of 13
Replace every (𝒙) by (𝟒𝒙𝟐 ) in the expansion of 𝟏⁄(𝟏 + 𝒙)
1
= 1 − (4𝑥 2 ) + (4𝑥 2 )2 − (4𝑥 2 )3 + ⋯ = 1 − 4𝑥 2 + 16𝑥 4 − 64𝑥 6 + ⋯
1 + 4𝑥 2

Differentiate the series for 𝟏⁄(𝟏 + 𝟒𝒙𝟐 ) with respect to 𝒙.

−8𝑥
2 2
= −8𝑥 + 64𝑥 3 − 384 𝑥 5 + ⋯
(1 + 4𝑥 )

Divide by −𝟖.
𝑥
∴ = 𝑥 − 8𝑥 3 + 48 𝑥 5 − ⋯
(1 + 4𝑥 2 )2

𝐞. 𝒇(𝒙) = 𝒙𝟐 𝐭𝐚𝐧−𝟏 (𝒙𝟑 )

1
Recall that = 1 − 𝑥 + 𝑥2 − 𝑥3 + ⋯ ⋯
1+𝑥
Replace every (𝒙) by (𝒙𝟐 ) in the expansion of 𝟏⁄(𝟏 + 𝒙)
1
2
= 1 − (𝑥 2 ) + (𝑥 2 )2 − (𝑥 2 )3 + (𝑥 2 )4 + ⋯ = 1 − 𝑥 2 + 𝑥 4 − 𝑥 6 + ⋯
1+𝑥

Integrate the series for 𝟏⁄(𝟏 + 𝒙𝟐 ) with respect to 𝒙.

−1
𝑥3 𝑥5 𝑥7
tan 𝑥 = 𝑥 − + − + ⋯+ 𝑐
3 5 7
Put 𝑥 = 0 in both sides, we get

tan−1 0 = 𝑐 ⟹ 𝑐=0

Replace every (𝒙) by (𝒙𝟑 ) in the expansion of 𝐭𝐚𝐧−𝟏 𝒙

−1 (𝑥 3 ) 3
(𝑥 3 )3 (𝑥 3 )5 (𝑥 3 )7 3
𝑥 9 𝑥 15 𝑥 21
tan =𝑥 − + − +⋯=𝑥 − + − +⋯
3 5 7 3 5 7

Multiply the series for 𝐭𝐚𝐧−𝟏 (𝒙𝟑 ) by 𝒙𝟐 .

2 −1 (𝑥 3 ) 5
𝑥 11 𝑥 17 𝑥 23
∴ 𝑥 tan =𝑥 − + − +⋯
3 5 7

Page 9 of 13
4. Evaluate each of the indefinite integrals as an infinite series:

cos 𝑥 − 1
a. ∫ 𝑥 cos(𝑥 3 ) 𝑑𝑥 c. ∫ 𝑑𝑥
𝑥

𝑒𝑥 − 1
b. ∫ 𝑑𝑥 d. ∫ tan−1 (𝑥 2 ) 𝑑𝑥
𝑥

Solution:

𝐚. ∫ 𝒙 𝐜𝐨𝐬(𝒙𝟑 ) 𝒅𝒙

1 2 1 4 1 6
Recall that cos 𝑥 = 1 − 𝑥 + 𝑥 − 𝑥 +⋯
2! 4! 6!

Replace every (𝒙) by (𝒙𝟑 ) in the expansion of 𝐜𝐨𝐬 𝒙

1 3 2 1 3 4 1 3 6 1 1 1
cos(𝑥 3 ) = 1 − (𝑥 ) + (𝑥 ) − (𝑥 ) + ⋯ = 1 − 𝑥 6 + 𝑥 12 − 𝑥 18 + ⋯
2! 4! 6! 2! 4! 6!

Multiply the series for 𝐜𝐨𝐬(𝒙𝟑 ) by 𝒙 .

1 7 1 13 1 19
𝑥 cos(𝑥 3 ) = 𝑥 − 𝑥 + 𝑥 − 𝑥 +⋯
2! 4! 6!
Integrate the both sides with respect to 𝒙.

3)
𝑥 2 𝑥 8 𝑥 14
∴ ∫ 𝑥 cos(𝑥 𝑑𝑥 = − + −⋯+𝑘
2 16 336

𝒆𝒙 − 𝟏
𝐛. ∫ 𝒅𝒙
𝒙
1 2 1 3
Recall that 𝑒𝑥 = 1 + 𝑥 + 𝑥 + 𝑥 +⋯⋯
2! 3!
Subtract 1 from the both sides.
1 2 1 3
𝑒𝑥 − 1 = 𝑥 + 𝑥 + 𝑥 + ⋯⋯
2! 3!

Page 10 of 13
Divide the series for 𝒆𝒙 − 𝟏 by 𝒙.

𝑒𝑥 − 1 1 1
= 1 + 𝑥 + 𝑥2 + ⋯ ⋯
𝑥 2! 3!
Integrate the both sides with respect to 𝒙.

𝑒𝑥 − 1 1 1
∴ ∫ 𝑑𝑥 = 𝑥 + 𝑥 2 + 𝑥 3 + ⋯ + 𝑐
𝑥 4 18

𝐜𝐨𝐬 𝒙 − 𝟏
𝐜. ∫ 𝒅𝒙
𝒙
1 2 1 4 1 6
Recall that cos 𝑥 = 1 − 𝑥 + 𝑥 − 𝑥 +⋯
2! 4! 6!
Subtract 1 from the both sides.
1 2 1 4 1 6
cos 𝑥 − 1 = − 𝑥 + 𝑥 − 𝑥 +⋯
2! 4! 6!
Divide the series for 𝐜𝐨𝐬 𝒙 − 𝟏 by 𝒙.

cos 𝑥 − 1 1 1 1
= − 𝑥 + 𝑥3 − 𝑥5 + ⋯
𝑥 2! 4! 6!
Integrate the both sides with respect to 𝒙.

cos 𝑥 − 1 −1 2 1 4 1
∴∫ 𝑑𝑥 = 𝑥 + 𝑥 − 𝑥6 + ⋯ + 𝑐
𝑥 4 96 4320

𝐝. ∫ 𝐭𝐚𝐧−𝟏 (𝒙𝟐 ) 𝒅𝒙

From the steps presented in problem 3 part (e), we got

−1
𝑥3 𝑥5 𝑥7
tan 𝑥 =𝑥− + − +⋯
3 5 7

Replace every (𝒙) by (𝒙𝟐 ) in the expansion of 𝐭𝐚𝐧−𝟏 𝒙

−1 (𝑥 2 ) 2
(𝑥 2 )3 (𝑥 2 )5 (𝑥 2 )7 2
𝑥 6 𝑥 10 𝑥 14
tan =𝑥 − + − +⋯=𝑥 − + − +⋯
3 5 7 3 5 7

Page 11 of 13
Integrate the both sides with respect to 𝒙.

1 1 1 1 15
∴ ∫ tan−1 (𝑥 2 ) 𝑑𝑥 = 𝑥 3 − 𝑥 7 + 𝑥 11 − 𝑥 + ⋯+ 𝑘
3 21 55 105

5. Find the second order Taylor series of each of the following functions:

a. 𝑓(𝑥) = √1 + 𝑥, 𝑎=2

b. 𝑓(𝑥) = 𝑥 cos 𝑥 , 𝑎=𝜋

c. 𝑓(𝑥) = tan−1 𝑥 , 𝑎=1

Solution:

𝐚. 𝒇(𝒙) = √𝟏 + 𝒙, 𝒂=𝟐

𝑓(𝑥) = √1 + 𝑥 = (1 + 𝑥)1/2 𝑓(2) = √3

1 1
𝑓 ′ (𝑥) = (1 + 𝑥)−1/2 𝑓 ′ (2) =
2 2 √3

−1 −1
𝑓 ′′ (𝑥) = (1 + 𝑥)−3/2 𝑓 ′′ (2) =
4 12 √3

−1
1 (12 )
√3
∴ √1 + 𝑥 = √3 + (𝑥 − 2) + (𝑥 − 2)2 + ⋯
2 √3 2!

𝐛. 𝒇(𝒙) = 𝒙 𝐜𝐨𝐬 𝒙 , 𝒂=𝝅

𝑓(𝑥) = 𝑥 cos 𝑥 𝑓(𝜋) = −𝜋

𝑓 ′ (𝑥) = cos 𝑥 − 𝑥 sin 𝑥 𝑓 ′ (𝜋) = −1

𝑓 ′′ (𝑥) = −2 sin 𝑥 − 𝑥 cos 𝑥 𝑓 ′′ (𝜋) = 𝜋

𝜋
∴ 𝑥 cos 𝑥 = −𝜋 − (𝑥 − 𝜋) + (𝑥 − 𝜋)2 + ⋯
2!
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𝐜. 𝒇(𝒙) = 𝐭𝐚𝐧−𝟏 𝒙 , 𝒂=𝟏

𝑓(𝑥) = tan−1 𝑥 𝑓(1) = 𝜋⁄4

1 1
𝑓 ′ (𝑥) = 𝑓 ′ (1) =
1 + 𝑥2 2

−2𝑥 −1
𝑓 ′′ (𝑥) = 𝑓 ′′ (1) =
(1 + 𝑥 2 )2 2

−1
𝜋 1 ( )
∴ tan−1 𝑥 = + (𝑥 − 1) + 2 (𝑥 − 1)2 + ⋯
4 2 2!

6. We know that a function 𝑦 = 𝑓(𝑥) has its tangent line approximation given by

𝑓(𝑥) ≅ 𝑓(𝑎) + 𝑓 ′ (𝑎) (𝑥 − 𝑎)

which is the first order Taylor series around 𝒂 = 𝟏. So,

𝑓(𝑥) ≅ 𝑓(1) + 𝑓 ′ (1) (𝑥 − 1)

∴ 𝑓(1.2) ≅ 3 − 2(1.2 − 1) = 3 − 2(0.2) = 2.6

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