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 1 st Derivative Intervals of Increasing and Decreasing Relative Extremas
Application of the 2 nd 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: lim x 2 3x2 + x + 5 =3( 2) 2 + 2 + 5 = 3 4 + 3 = 15
Special case Ex: lim x 2 x 2 4 x 2 = 0 0 Indeterminate form
Indeterminate Form Factoring lim x 2 x 2 4 x 2 Rationalizing (Conjugate) lim x 4 x 4 x 2 L'Hopital's Rule
Comparison Factoring lim x 2 x 2 4 x 2 =lim x 2 (x+2)(x 2) x 2 =lim x 2 x + 2 = 4 Rationalization lim x 4 x 4 x 2 =lim x 4 x 4 x 2 =lim x+2 x+2 (x 4)( x+2) x 4 x 4 =lim x 4 x + 2 = 4 conjugate a b a + b = a 2 + b 2
Comparison Factoring lim x 2 x 2 4 x 2 =lim x 2 (x+2)(x 2) x 2 =lim x 2 x + 2 = 4 L'Hopital's Rule lim x 2 x 2 4 x 2 = lim x 2 (x 2 4) (x 2) =lim x 2 2x 1 =2(2) = 4
Limits approaching Infinites For Polynomials Ex: lim x 5x 3 1 ALWAYS DNE!!! For Rational Functions lim x ± f(x) g(x) Three Cases Degree of TOP is Bigger Degree of BOTTOM is Bigger Same
Degree of Top > Degree of Bottom Case 1: The highest power is in the numerator. Always DNE (± ). Ex: lim 2x 5 x 3x lim x x 3 2 x 4 = DNE or = DNE or
Degree of Bottom > Degree of Top Case 2: The highest power is in the denominator Always = 0 Ex: lim x 4x 1 3x 4 +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: lim x 2x 3 +7x 4 5x 3 2x = 2 5
Solving for Discontinuities Goal: Turn a discontinued function into differentiable everywhere. Key: Sub in the restriction of x! 3x 4 if x 2 Ex: let x 2, what value of k + kx if x > 2 will make the function continuous everywhere? 3x 4 = x 2 + kx, while x = 2 3 2 4 = k = 1 2 2 + 2k
Derivatives 1. Power Rule x n 2. Product Rule f(x)g(x) 3. Quotient Rule f x g x 4. Chain Rule f x n
Marginal Analysis -It s the change analysis via the 1 st derivative!! C(x) -> Total Cost R(x) -> Total Revenue P(x) -> Total Profit C (x) -> Marginal Cost R (x) -> Marginal Revenue P (x) -> Marginal Profit Cost = Revenue Profit C(x) = R(x) P(x) R(x) = xf(x) f(x) is the price or demand function
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 x 2 R x = 300 2x Since 50 units were sold.. R 50 = 300 2(50) R 50 = $200 the approximate revenue of selling the 51 st 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 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 Demand is unitary If E(p)=1 -> same amount!!
Example Staples has determined that the demand for erasers is given by: P = 6 x. P is the unit price in 20 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 f x = 20 E 2 = 2 20 120 20 2 = 1 2 Inelastic. = Pf (p) f(p)
Implicit Differentiation y 3 + 7y = x 3 Two-Step Problem to find dy or dx y. 1) Differentiate both sides in respect of x. When differentiating y, put a y or dy after y dx 2) Solve the resulting function for y or dy in terms of x and y. dx
Let s solve it! y 3 + 7y = x 3 3y 2 y + 7(1)y = 3x 2 y 3y 2 + 7 = 3x 2 y = 3x2 3y 2 +7
1 st and 2 nd Derivative Application Increasing & Decreasing Intervals Local Extremas 1 st Derivative Slope Tangent
1 st and 2 nd Derivative Application Concavity Inflection Points 2 nd Derivative Point of Diminishing Return
Approach: Increasing/Decreasing Intervals 1. Take derivative (finding slopes of the function) f (x). 2. Set f 0 x = 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 3 x3 2x 2 6x 2, where is f(x) increasing/decreasing? f x = 2x 2 4x 6 0 = 2x 2 4x 6 0 = 2 x 2 2x 3 0 = x 3 x + 1 x = 3 1
Continued + + -1 3 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)
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
f x = x x 2 +1 max/min Quick Exercise, over 0,2 Find absolute f x = x2 +1 2x(x) x 2 +1 2 f x = 1 x2 x 2 +1 2 = 0 x = 1; x = 1 X=-1 not in the interval 0,2
1 x = 0 2 f x = x x 2 +1 f 1 = 1 2 f 0 = 0 Absolute Max Absolute Min f 2 = 2 5
Log/Ln Functions Since Log/Exponential Functions are inverses: 2 most important rules! e lnx = x, x > 0 lne x = x
Tricky Sample Exam Questions Solve for x: A) ln lnx = 1 e ln lnx = e 1 lnx = e e lnx = e e x = e e
Solve for x B) e ex = 10 lne ex = ln10 e x = ln10 lne x = ln ln10 x = ln ln10
Interests Problems General Compound Interest mt A = P 1 + r m Continuous Compound Interest A = Pe rt Effective Interest Rate (EFF) r eff = 1 + r 1 m Present Values General compounded (periodic) Continuous compounded (e) m
Effective Interest Rates Ex. Find the effective rate of interest corresponding to: a) 10%/year compounded semi-annually b) 9%/year compounded quarterly a) r eff = 1 + 0.10 2 b) r eff = 1 + 0.90 4 2 1 = 10.25%/year 4 1 = 9.308%/year
Summing up A = P A = Pe rt 1 + r m mt r eff = 1 + r m m 1 P = A 1 + r m P = Ae rt mt (Present Value) (Present Value)
TIME TO GET REAL
2008 Fall Exam 13. Find the derivative of f t = e 2t ln (t + 1). a) 2e2t t+1 b) e2t t+1 + 2e2t ln t + 1 c) 2e 2t + 1 t+1 d) e2t t+1 + e2t ln (t + 1)
2008 Fall Exam 10. Simplify ln 2e 5x + ln e3x 2 a)8x b)e 8x c)ln2 ln 1 2 + 8x d) e 5x 2 + e 3x 1 2
2008 Fall Exam 14. Find the derivative of f x = ln x 2 4 a) b) c) d) 2x x 2 4 x x 2 4 2x x 2 4 1 x 2 4
2008 Fall Exam 19. Find the equation of the tangent line to the graph of y = lnx at the x point 1,0. a) y = x + 1 b) y = x + 1 c) y = x 1 d) y = lnx x 1
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