Lecture 1 in calculus 1 math 122

FUNCTION:
Y=f(x)
Evaluation of a function
πΆπ‘œπ‘šπ‘π‘’π‘‘π‘’ π‘‘β„Žπ‘’ π‘–π‘›π‘‘π‘–π‘π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ π‘‘β„Žπ‘’ 𝑔𝑖𝑣𝑒𝑛 π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘›.
a) 𝑓(π‘₯) = 3π‘₯ + 5 π‘œπ‘Ÿ 𝑦 = 3π‘₯ + 1, 𝑓(0), 𝑓(1), 𝑓(2).
Answer: 𝑓(0) = 5, 𝑓(1) = 8, 𝑓(2) = 11
3 𝑖𝑓 𝑑 5
π‘Šβ„Žπ‘’π‘› 𝑑 = βˆ’6,
𝑓(βˆ’6) = 3, π‘€β„Žπ‘’π‘› 𝑑 = βˆ’11, 𝑓(βˆ’11) = 3
π‘Šβ„Žπ‘’π‘› 𝑑 = βˆ’1, 𝑓(βˆ’1) = 0, π‘€β„Žπ‘’π‘› 𝑑 = 0, 𝑓(0) = 1,
π‘Šβ„Žπ‘’π‘› 𝑑 = 16, 𝑓(16) = 4
CONSUMER DEMAND PROBLEMType equation here.
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘’ π‘‘π‘’π‘šπ‘Žπ‘›π‘‘ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘› 𝑝
= 𝐷(π‘₯) π‘Žπ‘›π‘‘ π‘‘β„Žπ‘’ π‘‘π‘œπ‘‘π‘Žπ‘™ π‘π‘œπ‘ π‘‘ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘› 𝐢(π‘₯)π‘“π‘œπ‘Ÿ π‘Ž π‘π‘Žπ‘Ÿπ‘‘π‘–π‘π‘’π‘™π‘Žπ‘Ÿ π‘π‘œπ‘šπ‘œπ‘‘π‘–π‘‘π‘¦ π‘Žπ‘Ÿπ‘’ 𝑔𝑖𝑣𝑒𝑛 𝑖𝑛 π‘‘π‘’π‘Ÿπ‘šπ‘  π‘œπ‘“ π‘‘β„Žπ‘’ 𝑙𝑒𝑣𝑒𝑙 π‘œπ‘“ π‘π‘Ÿπ‘œπ‘‘π‘’
𝐼𝑛 π‘’π‘Žπ‘β„Ž π‘π‘Žπ‘ π‘’, 𝑓𝑖𝑛𝑑
(π‘Ž) π‘‘β„Žπ‘’ π‘Ÿπ‘’π‘£π‘’π‘›π‘’π‘’ 𝑅(π‘₯) π‘Žπ‘›π‘‘ π‘π‘Ÿπ‘œπ‘“π‘–π‘‘ 𝑃(π‘₯)
(𝑏)𝐴𝑙𝑙 π‘£π‘Žπ‘™π‘’π‘’π‘  π‘œπ‘“ π‘₯π‘“π‘œπ‘Ÿ π‘€β„Žπ‘–π‘β„Ž π‘‘β„Žπ‘’ π‘π‘Ÿπ‘œπ‘‘π‘’π‘π‘‘π‘–π‘œπ‘› π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘œπ‘šπ‘œπ‘‘π‘–π‘‘π‘¦ 𝑖𝑠 π‘π‘Ÿπ‘œπ‘“π‘–π‘‘π‘Žπ‘π‘™π‘’.
π‘ƒπ‘Ÿπ‘œπ‘π‘™π‘’π‘š 1: 𝐷(π‘₯) = βˆ’0.02π‘₯ + 29
𝐢(π‘₯) = 1.43π‘₯ 2 + 18.3π‘₯ + 15.6
π‘€β„Žπ‘’π‘› π‘₯ = 5, 𝐷(5) = 28.99,
𝐢(5) = 142.85
𝐸π‘₯π‘Žπ‘šπ‘π‘™π‘’. 𝐴 π‘‘π‘–π‘“π‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ π‘žπ‘’π‘œπ‘‘π‘–π‘’π‘›π‘‘ 𝑖𝑠 π‘Žπ‘› 𝑒π‘₯π‘π‘Ÿπ‘’π‘ π‘ π‘–π‘œπ‘› 𝑖𝑛 π‘‘β„Žπ‘’ π‘“π‘œπ‘Ÿπ‘š π‘œπ‘“
𝑓(π‘₯ + β„Ž) βˆ’ 𝑓(π‘₯)
β„Ž
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘’ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘›
1. 𝑓(π‘₯) = π‘₯ + 1
π‘†π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘› π‘“π‘œπ‘Ÿ π‘›π‘’π‘šπ‘π‘’π‘Ÿ 2.
2. 𝑓(π‘₯) = π‘₯ 2 βˆ’ 3π‘₯
3. 𝑓(π‘₯) =
1
π‘₯
𝑓(π‘₯ + β„Ž) βˆ’ 𝑓(π‘₯)
=?
β„Ž
𝑓𝑖𝑛𝑑 π‘“π‘–π‘Ÿπ‘ π‘‘ 𝑓(π‘₯ + β„Ž) =?
𝑓(π‘₯) = π‘₯ 2 βˆ’ 3π‘₯
𝑓(π‘₯ + β„Ž) = (π‘₯ + β„Ž)2 βˆ’ 3(π‘₯ + β„Ž)
= π‘₯ 2 + 2β„Žπ‘₯ + β„Ž2 βˆ’ 3(π‘₯ + β„Ž)
𝑓(π‘₯ + β„Ž) = π‘₯ 2 + 2β„Žπ‘₯ + β„Ž2 βˆ’ 3π‘₯ βˆ’ 3β„Ž
𝑓(π‘₯ + β„Ž) βˆ’ 𝑓(π‘₯) (π‘₯ 2 + 2β„Žπ‘₯ + β„Ž2 βˆ’ 3π‘₯ βˆ’ 3β„Ž) βˆ’ (π‘₯ 2 βˆ’ 3π‘₯)
=
β„Ž
β„Ž
π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦,
𝑓(π‘₯ + β„Ž) βˆ’ 𝑓(π‘₯) π‘₯ 2 + 2β„Žπ‘₯ + β„Ž2 βˆ’ 3π‘₯ βˆ’ 3β„Ž βˆ’ π‘₯ 2 + 3π‘₯
=
β„Ž
β„Ž
𝑓(π‘₯ + β„Ž) βˆ’ 𝑓(π‘₯) 2β„Žπ‘₯ + β„Ž2 βˆ’ 3β„Ž β„Ž(2π‘₯ + β„Ž βˆ’ 3)
=
=
= 2π‘₯ + β„Ž βˆ’ 3 π‘“π‘–π‘›π‘Žπ‘™ π‘Žπ‘›π‘ π‘€π‘’π‘Ÿ.
β„Ž
β„Ž
β„Ž
Number 1 solution
𝑓(π‘₯) = π‘₯ + 1
π‘†π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
𝑓(π‘₯+β„Ž)βˆ’π‘“(π‘₯)
β„Ž
=
π‘₯+β„Ž+1βˆ’π‘₯βˆ’1 β„Ž
=β„Ž=1
β„Ž
Number 3 Answer:
𝑓(π‘₯ + β„Ž) βˆ’ 𝑓(π‘₯)
1
= βˆ’ 2,β„Ž = 0
β„Ž
π‘₯
𝐼𝑛𝑑𝑒𝑖𝑑𝑖𝑣𝑒 πΌπ‘›π‘‘π‘Ÿπ‘œπ‘‘π‘’π‘π‘‘π‘–π‘œπ‘› π‘‘π‘œ π‘‘β„Žπ‘’ π‘™π‘–π‘šπ‘–π‘‘
πΆπ‘œπ‘›π‘ π‘–π‘‘π‘’π‘Ÿ π‘Ž π‘šπ‘Žπ‘›π‘Žπ‘”π‘’π‘Ÿ π‘€β„Žπ‘œ π‘‘π‘’π‘‘π‘’π‘Ÿπ‘šπ‘–π‘›π‘’ π‘‘β„Žπ‘Žπ‘‘ π‘€β„Žπ‘’π‘› π‘₯ π‘π‘’π‘Ÿπ‘π‘’π‘›π‘‘ π‘œπ‘“ β„Žπ‘’π‘Ÿ π‘π‘œπ‘šπ‘π‘Žπ‘›π‘¦β€²π‘  π‘π‘™π‘Žπ‘›π‘‘ π‘π‘Žπ‘π‘Žπ‘π‘–π‘‘π‘¦ 𝑖𝑠
𝑏𝑒𝑖𝑛𝑔 𝑒𝑠𝑒𝑑, π‘‘β„Žπ‘’ π‘‘π‘œπ‘‘π‘Žπ‘™ π‘π‘œπ‘ π‘‘ π‘œπ‘“ π‘œπ‘π‘’π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘› 𝑖𝑠 𝐢 β„Žπ‘’π‘›π‘‘π‘Ÿπ‘’π‘‘ π‘‘β„Žπ‘œπ‘’π‘ π‘Žπ‘›π‘‘ π‘π‘’π‘ π‘œπ‘ , π‘€β„Žπ‘’π‘Ÿπ‘’
8π‘₯ 2 βˆ’ 636π‘₯ βˆ’ 320
𝐢(π‘₯) = 2
π‘₯ βˆ’ 68π‘₯ βˆ’ 960
π‘‡β„Žπ‘’ π‘π‘œπ‘šπ‘π‘Žπ‘›π‘¦ β„Žπ‘Žπ‘  π‘Ž π‘π‘œπ‘™π‘–π‘π‘¦ π‘œπ‘“ π‘Ÿπ‘œπ‘‘π‘Žπ‘‘π‘–π‘›π‘” π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘›π‘Žπ‘›π‘π‘’ 𝑖𝑛 π‘Žπ‘› π‘Žπ‘‘π‘‘π‘’π‘šπ‘ π‘‘π‘œ π‘’π‘›π‘ π‘’π‘Ÿπ‘’ π‘‘β„Žπ‘Žπ‘‘ π‘Žπ‘π‘π‘Ÿπ‘œπ‘₯π‘–π‘šπ‘Žπ‘‘π‘’π‘™π‘¦
80 π‘π‘’π‘Ÿπ‘π‘’π‘›π‘‘ π‘œπ‘“ π‘π‘Žπ‘π‘Žπ‘π‘–π‘‘π‘¦ 𝑖𝑠 π‘Žπ‘™π‘€π‘Žπ‘¦π‘  𝑖𝑛 𝑒𝑠𝑒. π‘Šβ„Žπ‘Žπ‘‘ π‘π‘œπ‘ π‘‘ π‘ β„Žπ‘œπ‘’π‘™π‘‘ π‘‘β„Žπ‘’ π‘šπ‘Žπ‘›π‘Žπ‘”π‘’π‘Ÿ 𝑒π‘₯𝑝𝑒𝑐𝑑 π‘€β„Žπ‘’π‘› π‘‘β„Žπ‘’
π‘π‘™π‘Žπ‘›π‘‘ 𝑖𝑠 π‘œπ‘π‘’π‘Ÿπ‘Žπ‘‘π‘–π‘›π‘” 𝑖𝑛 π‘‘β„Žπ‘–π‘  π‘–π‘‘π‘’π‘Žπ‘™ π‘π‘Žπ‘π‘Žπ‘π‘–π‘‘π‘¦?
𝐼𝑑 π‘šπ‘Žπ‘¦ π‘ π‘’π‘’π‘š π‘‘β„Žπ‘Žπ‘‘ 𝑀𝑒 π‘π‘Žπ‘› π‘Žπ‘›π‘ π‘€π‘’π‘Ÿ π‘‘β„Žπ‘’ π‘žπ‘’π‘’π‘ π‘‘π‘–π‘œπ‘› 𝑏𝑦 π‘ π‘–π‘šπ‘π‘™π‘¦ π‘’π‘£π‘Žπ‘™π‘’π‘Žπ‘‘π‘–π‘›π‘” 𝐢(80), 𝑏𝑒𝑑 π‘€β„Žπ‘’π‘› π‘₯
= 80, π‘‘β„Žπ‘’ π‘Ÿπ‘’π‘ π‘’π‘™π‘‘ 𝑖𝑠 π‘šπ‘’π‘Žπ‘›π‘–π‘›π‘”π‘™π‘’π‘ π‘ , π‘‘β„Žπ‘Žπ‘‘ 𝑖𝑠 π‘‘β„Žπ‘’ π‘Ÿπ‘’π‘ π‘’π‘™π‘‘ 𝑖𝑠 𝑒𝑛𝑑𝑒𝑓𝑖𝑛𝑑 π‘œπ‘Ÿ
0
π»π‘œπ‘€π‘’π‘£π‘’π‘Ÿ 𝑖𝑑 𝑖𝑠 𝑠𝑑𝑖𝑙𝑙 π‘π‘œπ‘ π‘ π‘–π‘π‘™π‘’ π‘‘π‘œ
0
π‘’π‘£π‘Žπ‘™π‘’π‘Žπ‘‘π‘’ 𝐢(π‘₯) π‘“π‘œπ‘Ÿ π‘£π‘Žπ‘™π‘’π‘’π‘  π‘œπ‘“ π‘₯ π‘Žπ‘π‘π‘Ÿπ‘‘π‘œπ‘Žπ‘β„Žπ‘’π‘  π‘‘π‘œ π‘₯
= 80. πΉπ‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘Ÿπ‘–π‘”β„Žπ‘‘ π‘œπ‘“ π‘₯ π‘Žπ‘›π‘‘ π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ 𝑙𝑒𝑓𝑑 π‘œπ‘“ π‘₯.
𝑋 π‘Žπ‘π‘π‘Ÿπ‘œπ‘Žπ‘β„Žπ‘’π‘  π‘‘π‘œ 80 π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ 𝑙𝑒𝑓𝑑
X
79.8
C(X)
6.99782
79.99
79.999
X approaches to 80 from the π‘Ÿπ‘–π‘”β„Žπ‘‘
80
80.0001
80.001
6.99989 6.99999 Undefined 7.000001 7.00001 7.00043
π‘‡β„Žπ‘’ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘›π‘Žπ‘™ π‘π‘’β„Žπ‘Žπ‘£π‘–π‘œπ‘Ÿ π‘œπ‘“ π‘‘β„Žπ‘–π‘  π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘› 𝐢(π‘₯) 𝑖𝑠 π‘‘π‘’π‘ π‘π‘Ÿπ‘–π‘π‘’π‘‘ 𝑏𝑦 π‘ π‘Žπ‘¦π‘–π‘› 𝑔
𝐢(π‘₯) π‘Žπ‘π‘π‘Ÿπ‘œπ‘Žπ‘β„Žπ‘’π‘  π‘‘π‘œ 𝑠𝑒𝑣𝑒𝑛 π‘Žπ‘  π‘₯ π‘Žπ‘π‘π‘Ÿπ‘œπ‘Žπ‘β„Žπ‘’π‘  π‘‘π‘œ 80
𝐼𝑛 π‘ π‘¦π‘šπ‘π‘œπ‘™, 𝑀𝑒 π‘π‘Žπ‘› π‘€π‘Ÿπ‘–π‘‘π‘’ π‘Žπ‘  lim 𝐢(π‘₯) = lim (
π‘₯β†’80
π‘₯β†’80
π·π‘’π‘“π‘–π‘›π‘–π‘‘π‘–π‘œπ‘› π‘œπ‘“ π‘™π‘–π‘šπ‘–π‘‘π‘  π‘Žπ‘›π‘‘ π‘π‘Ÿπ‘œπ‘π‘’π‘Ÿπ‘‘π‘–π‘’π‘  π‘œπ‘“ π‘™π‘–π‘šπ‘–π‘‘π‘ 
limπ‘₯→𝑐 𝑓(π‘₯) = 𝐿 π‘‘β„Žπ‘Žπ‘‘ 𝑖𝑠 limβˆ’ 𝑓(π‘₯) = 𝐿, π‘Žπ‘›π‘‘
π‘₯→𝑐
lim 𝑓(π‘₯) = 𝐿.
π‘₯β†’π‘₯ 𝑐
π‘ƒπ‘Ÿπ‘œπ‘π‘’π‘Ÿπ‘‘π‘–π‘’π‘  π‘œπ‘“ π‘™π‘–π‘šπ‘–π‘‘π‘ .
𝐸π‘₯π‘Žπ‘šπ‘π‘™π‘’. 𝐹𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘™π‘–π‘šπ‘–π‘‘ π‘œπ‘“ π‘‘β„Žπ‘’ π‘“π‘œπ‘™π‘™π‘œπ‘€π‘–π‘›π‘” π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘›π‘ 
1. lim π‘₯ 2 + 2π‘₯ =? π΄π‘›π‘ π‘€π‘’π‘Ÿ: lim (π‘₯ 2 + 2π‘₯) = 12 + 2(1) = 3
π‘₯β†’1
80.04
π‘₯β†’1
)=7
Remark
C(x)
approaches
to seven
2. lim (5π‘₯ + 3) =? π΄π‘›π‘ π‘€π‘’π‘Ÿ lim (5π‘₯ + 3) = lim 5π‘₯ + lim 3 = 5(2) + 3 = 13
π‘₯β†’2
⚫
π‘₯β†’2
π‘₯β†’2
π‘₯β†’2
The limit of constant is the given constant
lim π‘˜ = π‘˜
π‘₯→𝑐
𝐸π‘₯π‘Žπ‘šπ‘π‘™π‘’
⚫
lim 3 = 3
π‘₯β†’2
lim π‘˜π‘“(π‘₯) = π‘˜ lim 𝑓(π‘₯)
π‘₯→𝑐
π‘₯→𝑐
Find the limit: π₯𝐒𝐦 πŸπ’™ = 𝟐 π₯𝐒𝐦 𝒙 = 𝟐(𝟏) = 𝟐
π’™β†’πŸ
⚫
π’™β†’πŸ
𝑻𝒉𝒆 π’π’Šπ’Žπ’Šπ’• 𝒐𝒇 π’’π’–π’π’•π’Šπ’†π’π’• π’Šπ’” 𝒆𝒒𝒖𝒂𝒍 𝒕𝒐 𝒕𝒉𝒆 π’’π’–π’π’•π’Šπ’†π’π’• 𝒐𝒇 π’π’Šπ’Žπ’Šπ’•π’”
𝒇(𝒙)
𝒇(𝒙) π₯𝐒𝐦
= 𝒙→𝒄
𝒙→𝒄 π’ˆ(𝒙)
π₯𝐒𝐦 π’ˆ(𝒙)
π₯𝐒𝐦
𝒙→𝒄
π‘¬π’™π’‚π’Žπ’‘π’π’† π‘­π’Šπ’π’… 𝒕𝒉𝒆 π’π’Šπ’Žπ’Šπ’• 𝒐𝒇
⚫
π₯𝐒𝐦 𝒙 = 𝒄
⚫
πΏπ‘–π‘šπ‘–π‘‘π‘  π‘Žπ‘‘ 𝑖𝑛𝑓𝑖𝑛𝑖𝑑𝑦
𝒙
β€²
𝒙+𝟏
𝒙
𝒙+𝟏
π’™β†’πŸ•
𝒙 β‰  βˆ’πŸ π₯𝐒𝐦
=
π₯𝐒𝐦 𝒙
π’™β†’πŸ•
π₯𝐒𝐦 𝒙+𝟏
π’™β†’πŸ•
=
πŸ•
πŸ–
π’‚π’π’”π’˜π’†π’“
𝒙→𝒄
π‘‘β„Žπ‘Žπ‘‘ 𝑖𝑠 ,
lim 𝑓(π‘₯) = 𝐿, 𝑖𝑓 π‘‘β„Žπ‘’ π‘™π‘–π‘šπ‘–π‘‘ 𝑒π‘₯𝑖𝑠𝑑𝑠 π‘Žπ‘›π‘‘
π‘₯β†’+∞
lim 𝑓(π‘₯) = 𝑀
π‘₯β†’βˆ’βˆž
𝐸π‘₯π‘Žπ‘šπ‘π‘™π‘’, 𝑓𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘™π‘–π‘šπ‘–π‘‘ π‘œπ‘“ π‘‘β„Žπ‘’ 𝑔𝑖𝑣𝑒𝑛 π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘›
1
=?
π‘₯
lim
π‘₯+∞
π‘Žπ‘›π‘‘
lim
π‘₯β†’βˆ’βˆž
1
=?
π‘₯
x
1
2
10
1000
100000
β†’ +∞
1
π‘₯
1
.5
.1
.0001
.000001
Remark
1
π‘₯
approaches
to zero or
simply zero
1
π‘₯
π‘₯β†’βˆž
Therefore, lim
1. lim
π‘˜
π‘₯β†’βˆž π‘₯
2. lim
π‘₯β†’βˆž
=0
=0
𝐴
π‘₯π‘˜
1
π‘₯
π‘₯β†’βˆž 2
1. lim
=0
=
1
∞2
=
1
∞
=0
π‘₯
lim (π‘₯+1) =?
π‘₯β†’βˆž
By direct substitution,
π‘₯
∞
∞
lim (
)=
= = ∞ . 𝐼𝑠 π‘‘β„Žπ‘’ π‘Žπ‘›π‘ π‘€π‘’π‘Ÿ 𝑖𝑛𝑓𝑖𝑛𝑖𝑑𝑦 π‘π‘œπ‘Ÿπ‘’π‘π‘‘ π‘œπ‘Ÿ π‘›π‘œπ‘‘ π‘π‘œπ‘Ÿπ‘Ÿπ‘’π‘π‘‘?
π‘₯β†’βˆž π‘₯ + 1
∞+1 ∞
Technique in evaluating a limit involving fraction at infinity
π‘₯
π‘₯
1
1
1
π‘₯
lim (
) = lim π‘₯ , π‘ π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ π‘‘β„Žπ‘’ π‘Ÿπ‘’π‘ π‘’π‘™π‘‘ 𝑖𝑠 lim
=
=
= lim (
)
1
1
1
π‘₯β†’βˆž π‘₯ + 1
π‘₯β†’βˆž π‘₯
π‘₯β†’βˆž
π‘₯β†’βˆž π‘₯ + 1
1
+
0
1+π‘₯ 1+∞
π‘₯+π‘₯
𝑓𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘–π‘›π‘‘π‘–π‘π‘Žπ‘‘π‘’π‘‘ π‘™π‘–π‘šπ‘–π‘‘π‘ 
1. lim (3π‘₯ 2 βˆ’ 5π‘₯ + 2) = 4
π‘₯β†’2
π‘†π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›:
lim (3π‘₯ 2 βˆ’ 5π‘₯ + 2) = 3(22 ) βˆ’ 5(2) + 2 = 12 βˆ’ 10 + 2 = 4
π‘₯β†’2
π‘₯+1 4
2. lim π‘₯+2=5
π‘₯β†’3
2π‘₯+1
2 +2π‘₯βˆ’7
3π‘₯
π‘₯β†’βˆž
lim
π‘†π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›:
=0
2π‘₯ 1
2 1
2π‘₯ + 1
2 + π‘₯2
π‘₯ + π‘₯2
π‘₯
lim
= lim 2
= lim
2
2 7
π‘₯β†’βˆž 3π‘₯ + 2π‘₯ βˆ’ 7
π‘₯β†’βˆž 3π‘₯
π‘₯β†’βˆž
2π‘₯ 7
3+ βˆ’ 2
+ 2βˆ’ 2
π‘₯ π‘₯
π‘₯2
π‘₯
π‘₯
2
1
0+0
∞ + ∞2
=
=
=0
2
7
3+βˆžβˆ’ 2 3+0βˆ’0
∞
Therefore, lim
2π‘₯+1
π‘₯β†’βˆž 3π‘₯ 2 +2π‘₯βˆ’7
= 0, the limit exists.
Next topic:
Given the function 𝑦 = 𝑓(π‘₯). π‘Šβ„Žπ‘’π‘› π‘‘β„Žπ‘’ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘› 𝑓(π‘₯)𝑖𝑠 π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘  π‘Žπ‘‘ π‘π‘’π‘Ÿπ‘‘π‘Žπ‘–π‘› π‘›π‘’π‘šπ‘π‘’π‘Ÿ “π‘Ž”?
Answer: The function 𝑓(π‘₯) is continuous at x = a if the following conditions are satisfied:
1. 𝑓(π‘Ž) 𝑖𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑑
2. lim 𝑓(π‘₯) exists
π‘₯β†’π‘Ž
3. lim 𝑓(π‘₯) =𝑓(π‘Ž)
π‘₯β†’π‘Ž
Example:
Determine if the following functions are continuous at certain value of x=a.
1. 𝑓(π‘₯) = π‘₯, π‘₯ = 2
2. 𝑓(π‘₯) = π‘₯ 2 βˆ’ 1, π‘₯ = βˆ’1
1
3. 𝑓(π‘₯) = π‘₯
π‘₯ =0
Solution: #1
Condition 1: f(2)=2
Condition 2: lim π‘₯=2
π‘₯β†’2
Condition 3: lim π‘₯=f(2)=2
π‘₯β†’2
Therefore, the function f(x) = x is continuous at x = 2 since the three conditions are satisfied.
Solution: # 3
Condition 1:
March 9, 2022
Recall:
𝑓(π‘₯ + β„Ž) βˆ’ 𝑓(π‘₯)
β„Ž
π‘†π‘’π‘π‘π‘œπ‘ π‘’ 𝑀𝑒 π‘‘π‘Žπ‘˜π‘’ π‘‘β„Žπ‘’ π‘™π‘–π‘šπ‘–π‘‘ π‘œπ‘“ π‘‘β„Žπ‘’ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘› π‘Žπ‘  β„Ž π‘Žπ‘π‘π‘Ÿπ‘œπ‘Žπ‘β„Žπ‘’π‘  π‘‘π‘œ 0, we have
lim
β„Žβ†’0
𝑓(π‘₯+β„Ž)βˆ’π‘“(π‘₯)
β„Ž
the limit of this function as x approaches 0 is what we call as the derivative
π‘œπ‘“ π‘‘β„Žπ‘’ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘› 𝑓(π‘₯)
In symbol we can write that as
𝑓′(π‘₯) = lim
β„Žβ†’0
𝑓(π‘₯ + β„Ž) βˆ’ 𝑓(π‘₯)
β„Ž
Note.
The formula above is the formula how to find the derivative of the function.
The derivative of the function is the rate of change of the dependent variable with respect to the
independent variable.π‘‚π‘‘β„Žπ‘’π‘Ÿ π‘ π‘¦π‘šπ‘π‘œπ‘™ π‘œπ‘“ π‘‘β„Žπ‘’ π‘‘π‘’π‘Ÿπ‘–π‘£π‘Žπ‘‘π‘–π‘£π‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘›.
1. 𝑓’(π‘₯) read as f prime of x meaning derivative of y or f(x)
𝑑𝑦
2. 𝑑π‘₯ π‘Ÿπ‘’π‘Žπ‘‘ π‘Žπ‘  𝑑𝑦 π‘œπ‘“ 𝑑π‘₯, π‘œπ‘Ÿ π‘Ÿπ‘’π‘Žπ‘‘ π‘Žπ‘  π‘‘π‘’π‘Ÿπ‘–π‘£π‘Žπ‘‘π‘–π‘£π‘’ π‘œπ‘“ 𝑦 π‘€π‘–π‘‘β„Ž π‘Ÿπ‘’π‘ π‘π‘’π‘π‘‘ π‘₯
Note: 𝑓’(π‘₯) =
𝑑𝑦
𝑑π‘₯
Velocity= the rate of change of displacement with respect to time
In formula, 𝑣 =
𝑑𝑠
,
𝑑𝑑
Differentiation is the process of finding the derivative of the function
Find the derivative of the following functions.
1. 𝑓(π‘₯) = π‘₯
2. 𝑓(π‘₯) = π‘₯ 2 +1
1
3. 𝑓(π‘₯) = π‘₯
Use the formula to find the derivative:
𝑓(π‘₯ + β„Ž) βˆ’ 𝑓(π‘₯)
β„Žβ†’0
β„Ž
𝑓′(π‘₯) = lim
Given 𝑓(π‘₯) =x, find f’(x)=?
𝑓′(π‘₯) = lim
β„Žβ†’0
π‘₯+β„Žβˆ’π‘₯
β„Ž
= lim = lim 1 = 1
β„Žβ†’0
β„Ž
β„Ž β„Žβ†’0
Therefore, 𝑓′(π‘₯) = 1, π‘‡β„Žπ‘’ π‘‘π‘’π‘Ÿπ‘–π‘£π‘Žπ‘‘π‘–π‘£π‘’ π‘œπ‘“ π‘‘β„Žπ‘’ 𝑔𝑖𝑣𝑒𝑛 π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘› 𝑓(π‘₯) = π‘₯ 𝑖𝑠 1.
Problem #2
Given:
𝑓(π‘₯) = π‘₯ 2 +1, find the derivative.
Solution:
𝑓′(π‘₯) = lim
β„Žβ†’0
𝑓(π‘₯ + β„Ž) βˆ’ 𝑓(π‘₯)
β„Ž
(π‘₯ + β„Ž)2 + 1 βˆ’ (π‘₯ 2 + 1)
β„Žβ†’0
β„Ž
𝑓′(π‘₯) = lim
π‘₯ 2 + 2β„Žπ‘₯ + β„Ž2 + 1 βˆ’ π‘₯ 2 βˆ’ 1
β„Žβ†’0
β„Ž
𝑓′(π‘₯) = lim
2β„Žπ‘₯ + β„Ž2
β„Žβ†’0
β„Ž
𝑓′(π‘₯) = lim
𝑓′(π‘₯) = lim
β„Žβ†’0
β„Ž(2π‘₯ + β„Ž)
= lim (2π‘₯ + β„Ž) = 2π‘₯ + 0 = 2π‘₯
β„Žβ†’0
β„Ž
π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’, 𝑔𝑖𝑣𝑒𝑛 𝑓(π‘₯) = π‘₯ 2 + 1, π‘‘β„Žπ‘’ π‘‘π‘’π‘Ÿπ‘–π‘£π‘Žπ‘‘π‘–π‘£π‘’ 𝑖𝑠 𝑓′(π‘₯) = 2π‘₯.
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