# Lecture 1 in calculus 1 math 122

FUNCTION:
Y=f(x)
Evaluation of a function
πΆππππ’π‘π π‘βπ πππππππ‘ππ π£πππ’π ππ π‘βπ πππ£ππ ππ’πππ‘πππ.
a) π(π₯) = 3π₯ + 5 ππ π¦ = 3π₯ + 1, π(0), π(1), π(2).
Answer: π(0) = 5, π(1) = 8, π(2) = 11
3 ππ π‘ 5
πβππ π‘ = β6,
π(β6) = 3, π€βππ π‘ = β11, π(β11) = 3
πβππ π‘ = β1, π(β1) = 0, π€βππ π‘ = 0, π(0) = 1,
πβππ π‘ = 16, π(16) = 4
CONSUMER DEMAND PROBLEMType equation here.
πΊππ£ππ π‘βπ ππππππ ππ’πππ‘πππ π
= π·(π₯) πππ π‘βπ π‘ππ‘ππ πππ π‘ ππ’πππ‘πππ πΆ(π₯)πππ π ππππ‘πππ’πππ πππππππ‘π¦ πππ πππ£ππ ππ π‘ππππ  ππ π‘βπ πππ£ππ ππ πππππ’
πΌπ πππβ πππ π, ππππ
(π) π‘βπ πππ£πππ’π π(π₯) πππ ππππππ‘ π(π₯)
(π)π΄ππ π£πππ’ππ  ππ π₯πππ π€βππβ π‘βπ πππππ’ππ‘πππ ππ π‘βπ πππππππ‘π¦ ππ  ππππππ‘ππππ.
πππππππ 1: π·(π₯) = β0.02π₯ + 29
πΆ(π₯) = 1.43π₯ 2 + 18.3π₯ + 15.6
π€βππ π₯ = 5, π·(5) = 28.99,
πΆ(5) = 142.85
πΈπ₯πππππ. π΄ ππππππππππ ππ’ππ‘ππππ‘ ππ  ππ ππ₯ππππ π πππ ππ π‘βπ ππππ ππ
π(π₯ + β) β π(π₯)
β
πΊππ£ππ π‘βπ ππ’πππ‘πππ
1. π(π₯) = π₯ + 1
ππππ’π‘πππ πππ ππ’ππππ 2.
2. π(π₯) = π₯ 2 β 3π₯
3. π(π₯) =
1
π₯
π(π₯ + β) β π(π₯)
=?
β
ππππ ππππ π‘ π(π₯ + β) =?
π(π₯) = π₯ 2 β 3π₯
π(π₯ + β) = (π₯ + β)2 β 3(π₯ + β)
= π₯ 2 + 2βπ₯ + β2 β 3(π₯ + β)
π(π₯ + β) = π₯ 2 + 2βπ₯ + β2 β 3π₯ β 3β
π(π₯ + β) β π(π₯) (π₯ 2 + 2βπ₯ + β2 β 3π₯ β 3β) β (π₯ 2 β 3π₯)
=
β
β
ππππππππ¦,
π(π₯ + β) β π(π₯) π₯ 2 + 2βπ₯ + β2 β 3π₯ β 3β β π₯ 2 + 3π₯
=
β
β
π(π₯ + β) β π(π₯) 2βπ₯ + β2 β 3β β(2π₯ + β β 3)
=
=
= 2π₯ + β β 3 πππππ πππ π€ππ.
β
β
β
Number 1 solution
π(π₯) = π₯ + 1
ππππ’π‘πππ
π(π₯+β)βπ(π₯)
β
=
π₯+β+1βπ₯β1 β
=β=1
β
π(π₯ + β) β π(π₯)
1
= β 2,β = 0
β
π₯
πΌππ‘π’ππ‘ππ£π πΌππ‘ππππ’ππ‘πππ π‘π π‘βπ πππππ‘
πΆπππ ππππ π πππππππ π€βπ πππ‘ππππππ π‘βππ‘ π€βππ π₯ πππππππ‘ ππ βππ πππππππ¦β²π  πππππ‘ πππππππ‘π¦ ππ
πππππ π’π ππ, π‘βπ π‘ππ‘ππ πππ π‘ ππ ππππππ‘πππ ππ  πΆ βπ’πππππ π‘βππ’π πππ πππ ππ , π€βπππ
8π₯ 2 β 636π₯ β 320
πΆ(π₯) = 2
π₯ β 68π₯ β 960
πβπ πππππππ¦ βππ  π ππππππ¦ ππ πππ‘ππ‘πππ πππππ‘ππππππ ππ ππ ππ‘π‘πππ π‘π πππ π’ππ π‘βππ‘ ππππππ₯ππππ‘πππ¦
80 πππππππ‘ ππ πππππππ‘π¦ ππ  πππ€ππ¦π  ππ π’π π. πβππ‘ πππ π‘ π βππ’ππ π‘βπ πππππππ ππ₯ππππ‘ π€βππ π‘βπ
πππππ‘ ππ  ππππππ‘πππ ππ π‘βππ  πππππ πππππππ‘π¦?
πΌπ‘ πππ¦ π πππ π‘βππ‘ π€π πππ πππ π€ππ π‘βπ ππ’ππ π‘πππ ππ¦ π πππππ¦ ππ£πππ’ππ‘πππ πΆ(80), ππ’π‘ π€βππ π₯
= 80, π‘βπ πππ π’ππ‘ ππ  ππππππππππ π , π‘βππ‘ ππ  π‘βπ πππ π’ππ‘ ππ  π’πππππππ ππ
0
π»ππ€ππ£ππ ππ‘ ππ  π π‘πππ πππ π ππππ π‘π
0
ππ£πππ’ππ‘π πΆ(π₯) πππ π£πππ’ππ  ππ π₯ πππππ‘πππβππ  π‘π π₯
= 80. πΉπππ π‘βπ πππβπ‘ ππ π₯ πππ ππππ π‘βπ ππππ‘ ππ π₯.
π πππππππβππ  π‘π 80 ππππ π‘βπ ππππ‘
X
79.8
C(X)
6.99782
79.99
79.999
X approaches to 80 from the πππβπ‘
80
80.0001
80.001
6.99989 6.99999 Undefined 7.000001 7.00001 7.00043
πβπ ππ’πππ‘πππππ ππβππ£πππ ππ π‘βππ  ππ’πππ‘πππ πΆ(π₯) ππ  πππ ππππππ ππ¦ π ππ¦ππ π
πΆ(π₯) πππππππβππ  π‘π π ππ£ππ ππ  π₯ πππππππβππ  π‘π 80
πΌπ π π¦ππππ, π€π πππ π€πππ‘π ππ  lim πΆ(π₯) = lim (
π₯β80
π₯β80
π·ππππππ‘πππ ππ πππππ‘π  πππ πππππππ‘πππ  ππ πππππ‘π
limπ₯βπ π(π₯) = πΏ π‘βππ‘ ππ  limβ π(π₯) = πΏ, πππ
π₯βπ
lim π(π₯) = πΏ.
π₯βπ₯ π
πππππππ‘πππ  ππ πππππ‘π .
πΈπ₯πππππ. πΉπππ π‘βπ πππππ‘ ππ π‘βπ ππππππ€πππ ππ’πππ‘ππππ
1. lim π₯ 2 + 2π₯ =? π΄ππ π€ππ: 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 lim (5π₯ + 3) = lim 5π₯ + lim 3 = 5(2) + 3 = 13
π₯β2
β«
π₯β2
π₯β2
π₯β2
The limit of constant is the given constant
lim π = π
π₯βπ
πΈπ₯πππππ
β«
lim 3 = 3
π₯β2
lim ππ(π₯) = π lim π(π₯)
π₯βπ
π₯βπ
Find the limit: π₯π’π¦ ππ = π π₯π’π¦ π = π(π) = π
πβπ
β«
πβπ
π»ππ πππππ ππ ππππππππ ππ πππππ ππ πππ ππππππππ ππ ππππππ
π(π)
π(π) π₯π’π¦
= πβπ
πβπ π(π)
π₯π’π¦ π(π)
π₯π’π¦
πβπ
π¬ππππππ π­πππ πππ πππππ ππ
β«
π₯π’π¦ π = π
β«
πΏππππ‘π  ππ‘ πππππππ‘π¦
π
β²
π+π
π
π+π
πβπ
π β  βπ π₯π’π¦
=
π₯π’π¦ π
πβπ
π₯π’π¦ π+π
πβπ
=
π
π
ππππππ
πβπ
π‘βππ‘ ππ  ,
lim π(π₯) = πΏ, ππ π‘βπ πππππ‘ ππ₯ππ π‘π  πππ
π₯β+β
lim π(π₯) = π
π₯βββ
πΈπ₯πππππ, ππππ π‘βπ πππππ‘ ππ π‘βπ πππ£ππ ππ’πππ‘πππ
1
=?
π₯
lim
π₯+β
πππ
lim
π₯βββ
1
=?
π₯
x
1
2
10
1000
100000
β +β
1
π₯
1
.5
.1
.0001
.000001
Remark
1
π₯
approaches
to zero or
simply zero
1
π₯
π₯ββ
Therefore, lim
1. lim
π
π₯ββ π₯
2. lim
π₯ββ
=0
=0
π΄
π₯π
1
π₯
π₯ββ 2
1. lim
=0
=
1
β2
=
1
β
=0
π₯
lim (π₯+1) =?
π₯ββ
By direct substitution,
π₯
β
β
lim (
)=
= = β . πΌπ  π‘βπ πππ π€ππ πππππππ‘π¦ ππππππ‘ ππ πππ‘ πππππππ‘?
π₯ββ π₯ + 1
β+1 β
Technique in evaluating a limit involving fraction at infinity
π₯
π₯
1
1
1
π₯
lim (
) = lim π₯ , π πππππππ¦ π‘βπ πππ π’ππ‘ ππ  lim
=
=
= lim (
)
1
1
1
π₯ββ π₯ + 1
π₯ββ π₯
π₯ββ
π₯ββ π₯ + 1
1
+
0
1+π₯ 1+β
π₯+π₯
ππππ π‘βπ πππππππ‘ππ πππππ‘π
1. lim (3π₯ 2 β 5π₯ + 2) = 4
π₯β2
ππππ’π‘πππ:
lim (3π₯ 2 β 5π₯ + 2) = 3(22 ) β 5(2) + 2 = 12 β 10 + 2 = 4
π₯β2
π₯+1 4
2. lim π₯+2=5
π₯β3
2π₯+1
2 +2π₯β7
3π₯
π₯ββ
lim
ππππ’π‘πππ:
=0
2π₯ 1
2 1
2π₯ + 1
2 + π₯2
π₯ + π₯2
π₯
lim
= lim 2
= lim
2
2 7
π₯ββ 3π₯ + 2π₯ β 7
π₯ββ 3π₯
π₯ββ
2π₯ 7
3+ β 2
+ 2β 2
π₯ π₯
π₯2
π₯
π₯
2
1
0+0
β + β2
=
=
=0
2
7
3+ββ 2 3+0β0
β
Therefore, lim
2π₯+1
π₯ββ 3π₯ 2 +2π₯β7
= 0, the limit exists.
Next topic:
Given the function π¦ = π(π₯). πβππ π‘βπ ππ’πππ‘πππ π(π₯)ππ  ππππ‘πππ’ππ’π  ππ‘ ππππ‘πππ ππ’ππππ “π”?
Answer: The function π(π₯) is continuous at x = a if the following conditions are satisfied:
1. π(π) ππ  πππππππ
2. lim π(π₯) exists
π₯βπ
3. lim π(π₯) =π(π)
π₯βπ
Example:
Determine if the following functions are continuous at certain value of x=a.
1. π(π₯) = π₯, π₯ = 2
2. π(π₯) = π₯ 2 β 1, π₯ = β1
1
3. π(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:
π(π₯ + β) β π(π₯)
β
ππ’ππππ π π€π π‘πππ π‘βπ πππππ‘ ππ π‘βπ ππ’πππ‘πππ ππ  β πππππππβππ  π‘π 0, we have
lim
ββ0
π(π₯+β)βπ(π₯)
β
the limit of this function as x approaches 0 is what we call as the derivative
ππ π‘βπ ππ’πππ‘πππ π(π₯)
In symbol we can write that as
πβ²(π₯) = lim
ββ0
π(π₯ + β) β π(π₯)
β
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ππ π π¦ππππ ππ π‘βπ πππππ£ππ‘ππ£π ππ π‘βπ ππ’πππ‘πππ.
1. πβ(π₯) read as f prime of x meaning derivative of y or f(x)
ππ¦
2. ππ₯ ππππ ππ  ππ¦ ππ ππ₯, ππ ππππ ππ  πππππ£ππ‘ππ£π ππ π¦ π€ππ‘β πππ ππππ‘ π₯
Note: πβ(π₯) =
ππ¦
ππ₯
Velocity= the rate of change of displacement with respect to time
In formula, π£ =
ππ
,
ππ‘
Differentiation is the process of finding the derivative of the function
Find the derivative of the following functions.
1. π(π₯) = π₯
2. π(π₯) = π₯ 2 +1
1
3. π(π₯) = π₯
Use the formula to find the derivative:
π(π₯ + β) β π(π₯)
ββ0
β
πβ²(π₯) = lim
Given π(π₯) =x, find fβ(x)=?
πβ²(π₯) = lim
ββ0
π₯+ββπ₯
β
= lim = lim 1 = 1
ββ0
β
β ββ0
Therefore, πβ²(π₯) = 1, πβπ πππππ£ππ‘ππ£π ππ π‘βπ πππ£ππ ππ’πππ‘πππ π(π₯) = π₯ ππ  1.
Problem #2
Given:
π(π₯) = π₯ 2 +1, find the derivative.
Solution:
πβ²(π₯) = lim
ββ0
π(π₯ + β) β π(π₯)
β
(π₯ + β)2 + 1 β (π₯ 2 + 1)
ββ0
β
πβ²(π₯) = lim
π₯ 2 + 2βπ₯ + β2 + 1 β π₯ 2 β 1
ββ0
β
πβ²(π₯) = lim
2βπ₯ + β2
ββ0
β
πβ²(π₯) = lim
πβ²(π₯) = lim
ββ0
β(2π₯ + β)
= lim (2π₯ + β) = 2π₯ + 0 = 2π₯
ββ0
β
πβπππππππ, πππ£ππ π(π₯) = π₯ 2 + 1, π‘βπ πππππ£ππ‘ππ£π ππ  πβ²(π₯) = 2π₯.
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