Key Concepts. Important Techniques. 1. Average rate of change slope of a secant line. You will need two points ( a, the formula: to find value

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1 AB Calculus Unit Review Key Concepts Average and Instantaneous Speed Definition of Limit Properties of Limits One-sided and Two-sided Limits Sandwic Teorem Limits as x ± End Beaviour Models Continuity at a point Continuous functions Composites Intermediate Value Teorem Average Rates of Cange Instantaneous Rates of Cange Tangent to a curve Normal to a curve Slope of a curve Important Tecniques. Average rate of cange slope of a secant line You will need two points ( a, f ( a )) and ( b, f ( b) ). Use te formula ( ) f ( a) f b b a to find value. Instantaneous rate of cange slope of tangent line You need only one point, ( a, f ( a) ) Use te formula ( + ) ( ) f a f a to find value. Once you ave a slope and you need te equation of te tangent line or normal line (slope te is negative reciprocal) use te point te line is tangent at, ( x, y ), and te slope,m, in te formula: y y m x x ( ) to determine te equation of te line. 4. To determine te equation of a orizontal asymptote: - Find te End Beaviour model - Let x ± (remember you can ave two different answers)

2 AP Specific Practice Question ( ). Let f be te function defined by f x xe x for all real numbers x. Write an equation of te orizontal asymptote for te grap of f. ( ). Let f be te function given by f x x 7x+ 6. a) Find te zeros of f. b) Write te equation of te tangent line to te grap at x. Wat is te domain of f ( x) ln x x. 4. Find te x-intercept of ( ) ( ) ( ) f x sin x sin x, 0 x π 5. Wat is te instantaneous rate of cange at of te function f given by x f ( x) x x? ( ) ln x 0 < x 6. If f ( x ) ten find x ln( ), < x 4 ( ) f x x 7. If a 0, ten x a x a 4 x a is 4 π π cos + cos 8. Wat is li m 0 9. Evaluate te it: x + 4. x 5 x 5 0. Suppose f ( x) ( x) + for all x and tat f () is undefined. Wat is f ( x)? x

3 AB 998 Question. Evaluate te it 4x 8x 8. x + 6x + 5 x

4 AB 000 Question

5 Unit review Solutions x x f x xe, and te orizontal asymptotes occur wen x ± we need only x e ceck tese conditions.. Since ( ) Wen x te denominator grow a lot faster tan te numerator tus y0 is te equation of te asymptote. Wen x te quotient disappears, giving us no orizontal asymptotes Tus y0 is te orizontal asymptote (wic is only on te rigt and side of grap). Using a graping calculator we ave; a) Via te remainde teorem try f ( ) f ( ) f ( ) f ( ) f ( ) f ( ) f ( ) f ( f () ( ) ( )( + 6) f x x x x ( )( )( x x x+ ) Zeros are : x, x, x,,,,,, 6, 6) terefore x- is a factor, using polynomial division. b) To find te equation of te tangent line we need at point (given (, ) and a slope. To find ( + ) ( f x f x te slope we need to find li m wen x ) ) m ( + ) ( ) ( + ) ( ) f x f x f f

6 ( ) ( x ) Now using y y m x x y 4 + y 4x+ 8 x. Domain occurs wen > 0. x Now wen x> tis is true Wen x<0 tis is true f { or 0, } D x x> x< x R 4. f ( x) ( x) ( x) sin sin f(x)0 wen sin ( ) 0 or sin ( ) 0 x wen x 0, π sin ( x ) wen x ( ) π x sin x Te x-intercepts are: 0, π, π 5. we need to find ( + ) ( ) f x f x li m wen x ( ) ( ) ( ) ( ) + f ( + ) f ( ) +

7 + 4+ [ ] ( + ) From te top part of te piecewise function f ( x) ln ( ) x From te bottom part of piecewise function f ( x) ln ( ) + x 4ln( ) Since bot te one sided its are not te same. Te it d.n.e. 7. x a x a x a x + a x a ( )( ) x a 4 4 x a x a x a + a 8. Note: We can evaluate tis it in our ead after te next capter. Tis question reminds you know double angle teorems and special trig its. π π π π π cos + cos cos cos( ) sin sin ( ) cos m li 0 cos( ) sin ( ) 0 sin ( ) sin ( )

8 9. x+ 4 x+ 4 x+ 4+ x 5 x 5 x + 4+ ( x + 4) x 5 x 5 x 5 x 5 x 5 6 ( x 5)( x+ 4+ ) x 5 ( x 5)( x+ 4+ ) x x x x ( x) + f ( x) ( x) ( ) ( ) f x x + x x x x f

9 AB 998 Question Solution: Note: tis is a velocity time grap. Te rate of cange gives us accelerations. a) Te acceleration is positive 0 < t < 5 and 45 < t < 50 because v() t is increasing on 0 t 5 and 45 t 50. b) Average acceleration ( 50) v( 0) v ft Te average acceleration is. 50 sec c) Note: we don t ave an equation of te function. So we pick te closest two points and calculate te slope troug tose points. Slope of secant line troug ( 5, 8) and ( 40, 75) m ft Te approximation is 5sec

10 AB 000 Question Solution: a) Since runner A is represented by te grap. We need to determine te equation of te velocity for te appropriate section of te grap. Since we require te velocity at t, te slope of te slanted line is equation of 0 v 0 0 m t, ten v( ) s v t 0, given us te Runner B is dictated by te formula, terefore we need v ( ) m b) Te acceleration is just te slope of te velocity equation for A, s m s Te acceleration is ( + ) ( + ) v( + ) v( ) m 49 s