Math 110 Final Exam General Review - · PDF fileImplicit Differentiation ... Visually In...
Transcript of Math 110 Final Exam General Review - · PDF fileImplicit Differentiation ... Visually In...
Math 110 Final Exam General Review
Edward Yu
Da Game Plan
• Solving Limits – Regular limits – Indeterminate Form – Approach Infinities – One sided limits/Discontinuity
• Derivatives – Power Rule – Product/Quotient Rule – Chain Rule
• Marginal Functions – MR, MC, MP, Elasticity
• Implicit Differentiation & Related Rates
• Application of the 1st Derivative – Intervals of Increasing and Decreasing – Relative Extremas
• Application of the 2nd Derivative – Concavity – Inflection Points
• Law of Diminishing Returns
– The second derivative test
• Optimization – a.k.a Absolute Max/Min – Word Problems!!!
• Exponential Functions • Logarithmic Functions
– Ln, e.
• Compound Interests – General Compound Interest – Continuous Compound Interest – Effective Interest Rate – Present Value
• General compounded (periodic) • Continuous compounded (e)
Evaluating Limits
• Polynomials -> just substitute the limit in.
Ex: lim𝑥→−2
3𝑥2 + 𝑥 + 5
=3(−2)2+ −2 + 5 = 3 4 + 3 = 15
“Special case”
• Ex: lim𝑥→2𝑥2−4
𝑥−2
• =0
0
Indeterminate form
Indeterminate Form
Factoring
• lim𝑥→2𝑥2−4
𝑥−2
Rationalizing (Conjugate)
• lim𝑥→4
𝑥−4
𝑥−2
L'Hopital's Rule L'Hopital's Rule
Comparison
Factoring
• lim𝑥→2𝑥2−4
𝑥−2
=lim𝑥→2(𝑥+2)(𝑥−2)
𝑥−2
=lim𝑥→2 𝑥 + 2 = 4
Rationalization
• lim𝑥→4
𝑥−4
𝑥−2
=lim𝑥→4
𝑥−4
𝑥−2×
𝑥+2
𝑥+2
=lim𝑥→4
(𝑥−4)( 𝑥+2)
𝑥−4
=lim𝑥→4
𝑥 + 2
= 4
conjugate
𝑎 − 𝑏 𝑎 + 𝑏 = 𝑎2 + 𝑏2 𝑎 − 𝑏 𝑎 + 𝑏 = 𝑎2 + 𝑏2
Comparison
Factoring
• lim𝑥→2𝑥2−4
𝑥−2
=lim𝑥→2(𝑥+2)(𝑥−2)
𝑥−2
=lim𝑥→2 𝑥 + 2 = 4
L'Hopital's Rule
• lim𝑥→2𝑥2−4
𝑥−2
= lim𝑥→2
(𝑥2−4)′
(𝑥 − 2)′
=lim𝑥→22𝑥
1
=2(2) = 4
Limits approaching Infinites
• For Polynomials – Ex: lim
𝑥→∞5𝑥3 − 1
– ALWAYS DNE!!!
• For Rational Functions
– lim𝑥→±∞
𝑓(𝑥)
𝑔(𝑥)
– Three Cases 𝐷𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑇𝑂𝑃 𝑖𝑠 𝐵𝑖𝑔𝑔𝑒𝑟
𝐷𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝐵𝑂𝑇𝑇𝑂𝑀 𝑖𝑠 𝐵𝑖𝑔𝑔𝑒𝑟𝑆𝑎𝑚𝑒
Degree of Top > Degree of Bottom
• Case 1: The highest power is in the numerator.
Always DNE (±∞).
• Ex: lim𝑥→∞
2𝑥5
3𝑥4 = 𝐷𝑁𝐸 𝑜𝑟 ∞
lim𝑥→∞
𝑥3
2 𝑥= 𝐷𝑁𝐸 𝑜𝑟 ∞
Degree of Bottom > Degree of Top
• Case 2: The highest power is in the denominator
Always = 0
• Ex: lim𝑥→∞
4𝑥−1
3𝑥4+2= 0
Degree of Bottom = Degree of Top
• Case 3: The highest power is same on the top and the bottom.
limit is the quotient of the coefficients.
• Ex: lim𝑥→∞
2𝑥3+7𝑥−4
5𝑥3−2𝑥=
2
5
Solving for Discontinuities
• Goal: Turn a discontinued function into differentiable everywhere.
• Key: Sub in the restriction of x!
• Ex: let 3𝑥 − 4 𝑖𝑓 𝑥 ≤ 2
𝑥2 + 𝑘𝑥 𝑖𝑓 𝑥 > 2 , what value of k
will make the function continuous everywhere?
• 3𝑥 − 4 = 𝑥2 + 𝑘𝑥,𝑤ℎ𝑖𝑙𝑒 𝑥 = 2 • 3 2 − 4 = 2 2 + 2𝑘 • 𝑘 = −1
Derivatives…
1. Power Rule 𝑥𝑛 2. Product Rule 𝑓(𝑥)𝑔(𝑥)
3. Quotient Rule 𝑓 𝑥
𝑔 𝑥
4. Chain Rule 𝑓 𝑥𝑛
Marginal Analysis
-It’s the change analysis via the 1st derivative!!
• C(x) -> Total Cost • R(x) -> Total Revenue
• P(x) -> Total Profit • Cost = Revenue – Profit • C(x) = R(x) – P(x) • R(x) = xf(x)
– f(x) is the price or demand function
• C’(x) -> Marginal Cost • R’(x) -> Marginal
Revenue • P’(x) -> Marginal
Profit
Marginal Revenue Question
• Suppose the demand for computer speakers is given by 𝑃 𝑥 = 300 − 𝑥, where x is the number of speakers, and 𝑝 𝑥 is the unit cost in dollars. What’s the marginal revenue if 50 units were sold?
Marginal Revenue <- Revenue <- P(x)
Solution…
𝑅(𝑥) = 𝑝(𝑥) ∙ 𝑥 𝑅 𝑥 = 300 − 𝑥 𝑥 𝑅 𝑥 = 300𝑥 − 𝑥2 𝑅′ 𝑥 = 300 − 2𝑥 Since 50 units were sold.. 𝑅′ 50 = 300 − 2(50) 𝑅′ 50 = $200 −𝑡ℎ𝑒 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒 𝑟𝑒𝑣𝑒𝑛𝑢𝑒 𝑜𝑓 𝑠𝑒𝑙𝑙𝑖𝑛𝑔 𝑡ℎ𝑒 51𝑠𝑡 𝑢𝑛𝑖𝑡 𝑖𝑠 $200
Elasticity
• Elasticity of demand:
𝐸 𝑝 = −𝑃𝑓′(𝑝)
𝑓(𝑝)
→Watch OUT!! In a typical demand function,
𝑃 = 𝑓 𝑥 Switch into,
𝑥 = 𝑓 𝑃
Cause & Effect Causes
• Demand is Elastic
• Demand is Inelastic
• Demand is unitary
Effects
• If E(p)>1 -> small % change in price causes greater % change in demand
• If E(p)<1 -> small% change in price causes a even smaller % change in demand If E(p)=1 -> same amount!!
Example
• Staples has determined that the demand for erasers is given by: 𝑃 = 6 −
𝑥
20. P is the unit price in
dollars and x is the number of erasers sold per day. What’s the price elasticity when the price is at $2 per piece? What does the number mean?
• Hint: rearrange and solve for x
• 𝑃 = 6 −𝑥
20
•𝑥
20= 6 − 𝑝
• 𝑥 = 120 − 20𝑝 = f(p)
• Remember 𝐸 𝑝 = −𝑃𝑓′(𝑝)
𝑓(𝑝)
• 𝑓′ 𝑥 = −20
• 𝐸 2 = −2 −20
120−20 2=
1
2
• Inelastic.
Implicit Differentiation
𝑦3 + 7𝑦 = 𝑥3
• Two-Step Problem to find 𝑑𝑦
𝑑𝑥𝑜𝑟 𝑦′.
• 1) Differentiate both sides in respect of x. When differentiating y, put a
𝑦’ or 𝑑𝑦
𝑑𝑥 after y
• 2) Solve the resulting function for
𝑦′or 𝑑𝑦
𝑑𝑥 in terms of x and y.
Let’s solve it!
• 𝑦3 + 7𝑦 = 𝑥3 • 3𝑦2𝑦′ + 7(1)𝑦′ = 3𝑥2 • 𝑦′ 3𝑦2 + 7 = 3𝑥2
• 𝑦′ =3𝑥2
3𝑦2+7
1st and 2nd Derivative Application
Local Extremas
Local Extremas
1st
Derivative
Increasing & Decreasing
Intervals
Increasing & Decreasing
Intervals
Slope Slope Tangent Tangent
1st and 2nd Derivative Application
2nd Derivative
Concavity Concavity Inflection
Points Inflection
Points
Point of Diminishing
Return
Point of Diminishing
Return
Approach: Increasing/Decreasing Intervals
1. Take derivative (finding slopes of the function) f’(x).
2. Set 𝑓′ 𝑥 = 0
𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
3. Create open intervals by these points. 4. Select a test point C in each interval
and determine the sign of f’(c)
5. Intervals + 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔− 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
Exercise
• 𝑓 𝑥 =2
3𝑥3 − 2𝑥2 − 6𝑥 − 2, where is f(x)
increasing/decreasing?
• 𝑓′ 𝑥 = 2𝑥2 − 4𝑥 − 6 • 0 = 2𝑥2 − 4𝑥 − 6 • 0 = 2 𝑥2 − 2𝑥 − 3 • 0 = 𝑥 − 3 𝑥 + 1
• 𝑥 = 3−1
Continued
• 𝑓′ 𝑥 = (𝑥 − 3)(𝑥 + 1) • 𝑓′ −2 = −2 − 3 −2 + 1 = 5 (positive) • 𝑓′ 0 = 0 − 3 0 + 1 = −3 (negative) • 𝑓′ 5 = 5 − 3 5 + 1 = 12 (positive) • Increase on −∞,−1 ∪ (3,∞) • Decrease on (−1,3)
+
-1 3
− +
Process for finding Relative Extremas
1. Find critical points. (Domain!) 2. Determine the sign of f’(x) to the
left and right on each critical point.
+ → − Local Max − → + Local Min + → + NOTHING! − → − NOTHING!
Concavities
– 𝑓′′ 𝑥 > 0, 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒 𝑖𝑠 𝑐𝑜𝑛𝑐𝑎𝑣𝑒 𝑢𝑝 – 𝑓′′ 𝑥 < 0, 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒 𝑖𝑠 𝑐𝑜𝑛𝑐𝑎𝑣𝑒 𝑑𝑜𝑤𝑛
Law of Diminishing Return
• -point beyond which there are smaller & smaller returns for each $ invested.
Application of the second derivative From concave up to concave down
Visually
In Economics In Math 110
Optimization
• Goal: Find absolute Max/Min over an interval 𝑎, 𝑏
1. Find the critical points in the interval 𝑎, 𝑏
2. Compute the value of f at each critical point and check endpoints f(a), f(b)
3. Largest – Absolute Max Smallest – Absolute Min
Quick Exercise
• 𝑓 𝑥 =𝑥
𝑥2+1, over 0,2 Find absolute
max/min
• 𝑓′ 𝑥 =𝑥2+1−2𝑥(𝑥)
𝑥2+1 2
• 𝑓′ 𝑥 =1−𝑥2
𝑥2+1 2 = 0
• 𝑥 = 1; 𝑥 = −1
X=-1 not in the interval 0,2
• 𝑥 = 102
• 𝑓 𝑥 =𝑥
𝑥2+1
• 𝑓 1 =1
2
• 𝑓 0 = 0
• 𝑓 2 =2
5
Absolute Min
Absolute Max
Log/Ln Functions
• Since Log/Exponential Functions are inverses:
• 2 most important rules!
𝑒𝑙𝑛𝑥 = 𝑥 , 𝑥 > 0
𝑙𝑛𝑒𝑥 = 𝑥
Tricky Sample Exam Questions
• Solve for x:
• A) ln lnx = 1
• 𝑒ln 𝑙𝑛𝑥 = 𝑒1 • 𝑙𝑛𝑥 = 𝑒
• 𝑒𝑙𝑛𝑥 = 𝑒𝑒 • 𝑥 = 𝑒𝑒
Solve for x
• B) 𝑒𝑒𝑥= 10
• 𝑙𝑛𝑒𝑒𝑥= 𝑙𝑛10
• 𝑒𝑥 = 𝑙𝑛10 • 𝑙𝑛𝑒𝑥 = ln 𝑙𝑛10 • 𝑥 = ln 𝑙𝑛10
Interests Problems
• General Compound Interest
𝐴 = 𝑃 1 +𝑟
𝑚
𝑚𝑡
• Continuous Compound Interest 𝐴 = 𝑃𝑒𝑟𝑡
• Effective Interest Rate (EFF)
𝑟𝑒𝑓𝑓 = 1 +𝑟
𝑚
𝑚
− 1
• Present Values – General compounded (periodic) – Continuous compounded (e)
Effective Interest Rates Ex.
• Find the effective rate of interest corresponding to: a) 10%/year compounded semi-annually b) 9%/year compounded quarterly
a) 𝑟𝑒𝑓𝑓 = 1 +0.10
2
2− 1 = 10.25%/𝑦𝑒𝑎𝑟
b) 𝑟𝑒𝑓𝑓 = 1 +0.90
4
4− 1 = 9.308%/𝑦𝑒𝑎𝑟
Summing up…
• 𝐴 = 𝑃 1 +𝑟
𝑚
𝑚𝑡
• 𝐴 = 𝑃𝑒𝑟𝑡
• 𝑟𝑒𝑓𝑓 = 1 +𝑟
𝑚
𝑚− 1
• 𝑃 = 𝐴 1 +𝑟
𝑚
−𝑚𝑡 (Present Value)
• 𝑃 = 𝐴𝑒−𝑟𝑡 (Present Value)
TIME TO GET REAL
2008 Fall Exam
• 13. Find the derivative of 𝑓 𝑡 = 𝑒2𝑡ln (𝑡 + 1).
• a) 2𝑒2𝑡
𝑡+1
• b) 𝑒2𝑡
𝑡+1+ 2𝑒2𝑡 ln 𝑡 + 1
• c) 2𝑒2𝑡 +1
𝑡+1
• d) 𝑒2𝑡
𝑡+1+ 𝑒2𝑡ln (𝑡 + 1)
2008 Fall Exam
• 10. Simplify ln 2𝑒5𝑥 + ln𝑒3𝑥
2
• a)8𝑥 • b)𝑒8𝑥
• c)𝑙𝑛2 − ln1
2+ 8𝑥
• d) 𝑒5𝑥 2 + 𝑒3𝑥1
2
2008 Fall Exam
• 14. Find the derivative of
𝑓 𝑥 = 𝑙𝑛 𝑥2 − 4
a)2𝑥
𝑥2−4
b)𝑥
𝑥2−4
c)2𝑥
𝑥2−4
d)1
𝑥2−4
2008 Fall Exam
• 19. Find the equation of the tangent line to the graph of 𝑦 =
𝑙𝑛𝑥
𝑥 at the
point 1,0 .
a) 𝑦 = 𝑥 + 1 b) 𝑦 = −𝑥 + 1 c) 𝑦 = 𝑥 − 1
d) 𝑦 =𝑙𝑛𝑥
𝑥− 1
The End