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AP Calculus AB AP Exam Review Limits Properties of Limits 1. Scalar Multiple: lim k f x k lim f x xc xc 2. Sum/Difference: lim f x g x xc lim f x lim g x xc xc lim f x g x lim f x xc 3. Product: xc f x lim xc g x 4. Quotient: lim f x xc lim g x xc lim g x xc if lim g x 0 xc Definition of Limit: as x gets closer and closer to a specific value, y gets closer and closer to what value? - recall, the answer to a limit is the y-value (but the y-value does not have to exist to have a limit value) One-Sided Limits lim f x means x approaches a from the right lim f x means x approaches a from the left x a x a Two-Sided Limit : lim f x means x approaches a from both sides xa Only exists if lim x a f x lim x a f x L , where L is a Real number Limits at Infinity If lim f x L or x Recall: To evaluate lim x Also: If lim x a lim x f x L , then y L f x g x , use is a horizontal asymptote highest degree term ( x if an x remains) highest degree term small BIG 0 and BIG small f x or lim f x , then x a is a vertical asymptote xa AP Calculus AB AP Exam Review Continuity i) lim f x xa lim x a f x ii) f a exists iii) lim f x xa (Roads meet) (Bridge is built) f a (Roads meet at the bridge) Types of Discontinuity Jump Removable Infinite 3x 2 x f x Example Problem: If 2x f 0 k , x 0 and if f is continuous at x 0 , then k (A) 3 2 (B) 1 (C) 0 (D) 1 (E) 3 2 Two Major Theorems on Continuity ( I ) Intermediate Value Theorem : If f is continuous on [ a , b ] and k is any number between f a and f b , then there is at least one number c between a and b such that f c k . ( II ) Differentiability Implies Continuity - If the derivative exists at a point, the function is continuous at that point AP Calculus AB AP Exam Review Differential Calculus Definition of Derivative (i) f ' a lim f a h f a h0 (ii) f ' a lim h f x f a x a xa (iii) f ' a lim f a h f a h h0 Symmetric Difference Quotient 2h (good to use given a table of values) Terminology for Derivative: Slope of the curve Instantaneous Rate of Change Slope of the tangent Rules Power Rule: d n x n x n 1 dx d f g f g ' g f ' dx Product Rule: (1st times derivative of 2nd plus 2nd times derivative of 1st) Quotient Rule: d f g f ' f g' dx g g2 (bottom times derivative of top minus top times derivative of bottom all over bottom squared) Chain Rule: d f g x f ' g x g ' x (derivative of outside at inside times derivative of dx inside) Trig Derivatives d du sin u cos u dx dx d du tan u sec 2 u dx dx d du sec u sec u tan u dx dx d du cos u sin u dx dx d du cot u csc 2 u dx dx d du csc u csc u cot u dx dx AP Calculus AB AP Exam Review Inverse Trig Derivatives d sin 1 u dx d cos 1 u dx 1 du 1 u 2 dx 1 du 1 u 2 dx d 1 du tan 1 u dx 1 u 2 dx d 1 du cot 1 u dx 1 u 2 dx d sec1 u dx u d 1 du cs c1 u 2 dx u u 1 dx 1 du u 1 dx 2 Exponential / Log Derivatives d u du e eu dx dx d du a u a u ln a dx dx d 1 du ln u dx u dx Implicit Differentiation - use when you cannot solve for y easily or solving for y makes taking the derivative more difficult - Remember: anytime you take a derivative of a function with y, dy you must put after that term dx dy - Then solve the resulting equation for dx ** Don’t forget about PRODUCT RULE on these problems! Example Problem: Find Answer: dy dx given y 3 xy 2 y x 2 2 . dy 2x y 2 dx 3y x 2 Recall also: We use Implicit Differentiation in Related Rates problems (differentiate variables w/ respect to t) AP Calculus AB AP Exam Review Derivatives of Inverse Functions If y f x and x f y are inverse functions, then the derivative of the inverse function is the reciprocal of the derivative of the original function: Alternative Notation: d f dx 1 evaluated at x f a = dx 1 dy dy dx 1 f 'a ** Recall: The specific value where you want to evaluate the derivative of the inverse function is the y-value of the original function, so you have to find the x-value on the original function which will give you that specific y-value and use this value to evaluate the derivative of the original function. Example Problem: (May use calculator) Let f x x 3 7 x 2 25x 39 and let g be the inverse of f (x). What is the value of g ' 0 ? (A) 1 25 Answer: C (B) 1 25 (C) 1 10 (D) 10 (E) 25 AP Calculus AB AP Exam Review Logarithmic Differentiation Use when function is very complicated OR if you have a function function Steps: 1. Take ln of both sides 2. Differentiate 3. Solve for dy dx 4. Substitute for y x 1 y 2 x 1 2 Example 1: 1 3 1 x2 1 3 ln y ln 2 x 1 1 x2 1 1 ln x 2 1 ln x 2 1 ln y ln 2 3 x 1 3 1 dy 1 1 1 2 2x 2 2 x y dx 3 x 1 x 1 dy 1 2x 2x 2 y 2 dx 3 x 1 x 1 dy 1 2x dx 3 x 2 1 Example 2: y tan x 1 2x x2 1 3 x2 1 x 2 1 x ln y ln tan x x ln y x ln tan x 1 dy 1 x sec 2 x ln tan x 1 y dx tan x dy 1 x sec 2 x ln tan x 1 y dx tan x dy 1 x sec 2 x ln tan x 1 dx tan x tan x x (can’t use power rule) AP Calculus AB AP Exam Review Equation of Tangent Line - Must have point = a , f a , and slope = f ' a - y f a f ' a x a Equation of Normal Line - Perpendicular to tangent at point of tangency, a , f a - y f a 1 x a (same equation with slope negative reciprocal) f 'a Linearization - using the y-value of the tangent line to approximate the function value Equation of tangent line y f a f ' a x a becomes L x f ' a x a f a ** Remember x = “ugly” number (one with the decimal) and a = “pretty number” (close to x) Mean Value Theorem for Derivatives If f is continuous on [a , b] and differentiable on (a , b), then there exists a number c between a and b such that f 'c - Slope of tangent line Instantaneous rate of change f b f a b a = = slope of secant line average rate of change Extreme Value Theorem If f is continuous on [a , b], then f has both an absolute maximum and an absolute minimum. - Max or min must occur at critical points f ' x 0 or f ' x DNE or at the endpoints. AP Calculus AB AP Exam Review Curve Sketching Rules f ' c 0 (horizontal tangent) Critical points occur when: f ' c DNE (corner, cusp, vertical tangent) (denominator = 0) Situation Indicates f ' x 0 f x is increasing f ' x 0 f x is decreasing f '' x 0 f x is concave up or f ' x is increasing f '' x 0 f x is concave down or f ' x is decreasing 1st Derivative Test – used to justify max and mins 0 and f ' c 0 f ' c 0 and f ' c 0 Max at x c if f ' c Min at x c if (positive to negative) (negative to positive) ** Must have a sign change to be a max or min, but it is still a critical point 2nd Derivative Test – used to justify max and mins Max at x c if f ' c 0 and f '' c 0 (horizontal tangent, concave down) Max at x c if f ' c 0 and f '' c 0 (horizontal tangent, concave up) Points of Inflection - where the graph changes concavity Example Problems: Use the graph of [sign change in f '' x ] y f ' x , sketched below on [0 , 7] 1. From the graph it follows that ______________. y f ' x A) f is discontinuous at x 4 B) f is decreasing for 4 x 7 C) f is constant for 0 x 4 D) f has a local maximum at x 0 E) f has a local minimum at x 7 2. Which best describes f at x 5 ? A) f has a root B) f is a maximum C) f is a minimum D) f has a point of inflection 3. A) B) C) D) The function is concave downward for which interval? (0 , 4) (4 , 5) (5 , 7) (4 , 7) AP Calculus AB AP Exam Review Optimization - Maximize or minimize a quantity - Write an equation that needs to be maximized or minimized (usually has 2 independent variables) - Write 2nd equation that relates the 2 independent variables, then substitute into 1st equation - Differentiate, find critical points, use sign chart to justify max/min, and check endpoints Related Rates – Implicit differentiation where you differentiate with respect to t - Write equation that relates variables Differentiate with respect to t Substitute known values (usually use 1st equation to find a missing value) Then solve for correct rate. Position, Velocity, Acceleration x t position of particle at time t v t x ' t instantaneous velocity at time t a t v ' t x '' t instantaneous acceleration at time t Speed = v t Speeding Up - velocity and acceleration have same sign Slowing Down - velocity and acceleration have opposite signs AP Calculus AB AP Exam Review Integral Calculus Definite Integrals: b a f x dx - represents the NET AREA under the curve (above x-axis is a positive value, below is a negative value) ** Answer is always a numerical value Indefinite Integrals: f x dx ** DON”T FORGET: +C - tells you to find ALL possible antiderivatives Techniques for Integration Power Rule in Reverse: u n du un 1 C (Add one to the exponent, n 1 divide by new exponent) Substitution Method: - “Chain Rule” for integrals - composite functions (function inside a function) f g x g ' x dx F g x C Steps: Let u g x So we have du g ' x dx Take derivative of u: Substitute du Take antiderivative Replace u with g x f u g ' x dx so f u du f u du F u C F g x C Separable Differential Equations dy dx Get y with dy and dx with x: Integrate and solve for y: 1 dy g y 1 g y dy f x g y f x dx f x dx y F x C AP Calculus AB AP Exam Review To Approximate Definite Integrals LRAM, RRAM, MRAM - rectangular approximation methods - approximate area under curve or value of a definite integral (usually use when given values in a table) Trapezoidal Rule : Recall: A trap 1 b1 b2 h ** must use this formula if values of x 2 are not uniform ** If x has same values throughout, you can use formula 1 f x 0 2 f x1 2 f x 3 ... 2 f x n 1 f x n x 2 A FUNDAMENTAL THEOREM OF CALCULUS Part I: Derivatives and Integrals are inverses d dx x a f t dt f x ** Remember, integral must start at a constant, and only have a variable as the upper limit ** Also, the independent variable for f becomes upper limit variable Part II (Evaluation): b a f x dx F b F a (antiderivative at upper limit minus antiderivative at lower limit) CHAIN RULE Version If y g x a f t dt , find dy . dx Steps: Evaluate integral using Part I Let F t be any antiderivative of f t , so Now find dy using chain rule, dx but since F ' f , g x a f t dt F g x F a , so y F g x F a dy F ' g x g ' x 0 , dx dy f g x g ' x dx AP Calculus AB AP Exam Review 1 b a avg f Average Value of a Function b a f x dx MEAN VALUE THEOREM FOR INTEGRALS - If f is continuous on [a , b], then at some point c in [a , b], 1 b a f c b a f x dx ** States that there is an actual y-value on f equals the average y-value of the function Position, Velocity, Acceleration Displacement: Net distance = Total Distance Traveled: b a b a v t dt v t dt (on a calculator) x b x a Final Position = Initial Position + Displacement or Area Between Curves y b y bottom dx a top b r 2 dx x d or right c b a v t dt x left dy Volumes of Revolution - Disk Method: (around x-axis) - Washer Method: b a a R 2 r 2 dx or or (around y-axis) d c b a d c d c R 2 r 2 dy Volumes of Known Cross-Sections A x dx (cross-sections perpendicular to x-axis) A y dy (cross-sections perpendicular to y-axis) r 2 dy