AP Calculus I Summer Packet
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1 AP Calculus I Summer Packet This will be your first grade of AP Calculus and due on the first day of class. Please turn in ALL of your work and the attached completed answer sheet. I. Intercepts The -intercept is where the graph crosses the -ais. You can find the -intercept by setting y=0. The y-intercept is where the graph crosses the y-ais. You can find the y-intercept by setting =0. Eample: Find the intercepts for y ( ) 4 Solution: -intercept 0 ( ) 4 set y=0 4 ( ) take square root of both sides ( ) or ( ) Write as equations 5 or Subtract from both sides ( ) add 4 to both sides y-intercept y (0 ) 4 set =0 Add 0+ y 4 y 9 4 Square y 5 subtract Problem Set I Find the intercepts for each of the following.. y 6 4. y 4 0. y 4. y 6 5. y 6
2 II. Lines The slope intercept form of a line is y = m + b where m is the slope and b is the y-intercept. The point slope form of a line is y y m ) where m is the slope and ) is a point on the line. In Calculus ( the point slope form of a line is the preferred form. If two lines are parallel then they have the same slope. If two lines are perpendicular then they have negative reciprocal slopes. ( y, Eample: Find the slope of the lines parallel and perpendicular to y 5 Solution: The slope of this line is m The parallel line has slope m and the perpendicular line has slope m. Eample: Find the equations of (a) line parallel and (b) perpendicular to y 5 that contains the point (-,) Solution: Part a (using slope from eample above) y = ( + ) Using the point-slope form with m and point (-,) Part b (using slope from eample above) y = ( + ) Using the point-slope form with m and point (-,) Eample: Find the slope and y-intercept of 6 5y 5 Solution: First you must get the line in slope-intercept form. 5y 5 6 Subtract 6 form both sides 5 6 Divide by -5 y 5 6 Simplify y 5 The slope is m= 6 and the y-intercept is - 5 Eample: Find the equation of the line that passes through (-,) and (4,5). Solution: You will need to find slope using m y y 5 m choose one point to substitute back into either the point slope or slope-intercept form of a 4 5 line. y 5 = ( 4) Using the point-slope form with m and point (4,5) 5 5
3 Problem Set II Find the equation of a line:. contains (,-4) and (5,). contains,, 4 6. contains (-,4) and m is undefined 4. contains (-,-) and m= 5. -intercept (,0) and y-intercept (0,) Find the slope and y-intercept of the line: 6. +5y = 0 Sketch a graph of the equation: 7. y=- 8. =4 9. y-=(+4) Write an equation of a line through the point (a). parallel to the given line and (b) perpendicular to the given line: 0. Point : (,) line: 4-y=
4 III. Functions Definition: Let f and g be functions. The function given by (f g)()=f(g()) is called the composite of f with g. The domain of f g is the set of all in the domain of g such that g() is in the domain of f. Eample: Given: f()=+5 and g()=- Find: f(g()), g(f()) and f(g()) Solution: To find f(g()) we must first find g(): g()=()- =4-= Since g()= we can find f(g())=f()=()+5=9+5=4 To find g(f()) we must first find f(): f()=()+5=6+5= Since f()= we can find g(f())=g()=()-=-= To find f(g()) we must put the function g() into f() equation in place of each. f(g())=f(-)=(-)+5=6-+5=6+ Eample: Given f() = + find f( + h). Solution: To find f( + h) we replace every instance of with +h. Thus f( + h) = ( + h) + + h = ( + h + h ) + + h = + 6h + h + + h The domain of a function is the set of values for which the function is defined. The range of a function is the set of y values that a function can return. In Calculus we usually write domains and ranges in interval notation. If the domain were -< 7 then in interval notation the domain would be (-,7]. Notice that the left side has a ( because it does not include - but the right side includes 7 so we use a ]. When using interval notation we never use a [ or ] for infinity. Eample: Find the domain and range for Solution: Since we can only take the square root of positive numbers - 0 which means that. So we would say the domain is [, ). Note that we have used a [ to indicate that is included. If was not to be included we would have used (, ). The smallest y value that the function can return is 0 so the range is (0, ). Problem Set III Let f() = + and g() = find each of the following:. g(f()). f( + 5). g( + h) g() Find the domain and range for each function give your answer using interval notation: 4. h()= 4 5. f()= For each function below, write functions f() and g() such that h() = f(g()) 6. h() = + 7. h() = sin
5 IV. Asymptotes and Holes Given a rational function if a number causes the denominator and the numerator to be 0 then both the numerator and denominator can be factored and the common zero can be cancelled out. This means there is a hole in the function at this point. Eample: Find the holes in the following function Solution: When = is substituted into the function the denominator and numerator both are 0. Factoring and canceling: ( )( ) but ( ) this restriction is from the original function before canceling. The graph of the ( ) function f() will look identical to y ecept for the hole at =. ( ) ( ) f note the hole at = y ( ) Given a rational function if a number causes the denominator to be 0 but not the numerator to be 0 then there is a vertical asymptote at that value. Eample: Find the vertical asymptotes for the function Solution: When =- is substituted into f() then the numerator is - and the denominator is 0 therefore there is an asymptote at =. See the graphs above. Given a rational function if a number causes the numerator to be 0 but not the denominator to be 0 then the value is an -intercept for the rational function. Eample: Discuss the zeroes in the numerator and denominator Solution: When =- is substituted into the function the numerator is 0 and the denominator is -6 so the value of the function is f(-)=0 and the graph crosses the -ais at =-. Also note that for =0 the numerator is and the denominator is 0 so there is a vertical asymptote at =0. The graph is below.
6 Eample: Find the holes, vertical asymptotes and -intercepts for the given function: f ) 6 ( Solution: First we must factor to find all the zeroes for both the numerator and denominator: ( ) ( ) Numerator has zeroes =0 and = Denominator has zeroes =0 and =-. =0 is a hole =- is a vertical asymptote = is a -intercept Also, recall horizontal asymptotes can be found by comparing the degree of the numerator and denominator of a rational function. If the denominator is of higher degree, the horizontal asymptote is y = 0. If the numerator is of higher degree, there is no horizontal asymptote. If the numerator and denominator are the same degree, the horizontal asymptote is y = k where k is the ratio of the lead coefficients. Eample: Find the horizontal asymptote of f() = Solution: Since the degrees of both polynomials is, we use the ratio of the lead coefficients to get an asymptote of y =. Problem Set IV For each function below list all holes, vertical asymptotes and -intercepts, and horizontal asymptotes.. ( )( ) ( )( ). y g ( )
7 V. Trig. Equations and Special Values You are epected to know the special values for trigonometric functions. Fill in the table below and study it. (degrees) (radians) cos sin Quadrant You should study the following trig identities and memorize them before school starts: Reciprocal identities sin cos csc sec csc sec sin cos tan cot cot tan Tangent Identities sin tan cos cos cot sin Pythagorean Identities sin cos tan sec cot csc Reduction Identities sin( ) sin cos( ) cos tan( ) tan We use these special values and identities to solve equations involving trig functions. Eample: Find all solutions to sin sin Solution: sin sin Original Problem
8 sin sin 0 ( sin )(sin ) 0 ( sin ) 0 and (sin ) 0 sin and sin k 6 and k 5 k 6 Get one side equal to 0. Factor Set each factor equal to 0 Problem Set V Find all solutions to the equations. You should not need a calculator.. 4cos 4cos. sin sin 0. sin cos Get the trig function by itself Solve for (these are special values)
9 VI. Eponents A fractional eponent means you are taking a root. For eample / is the same as. Eample: Write without fractional eponent: / y Solution: number in the fraction. y Notice that the root is the bottom number in the fraction and the power is the top Negative eponents mean that you need to take the reciprocal. For eample means. y 5 Eample: Write with positive eponents: 4 Solution: y 4 5 Eample: Write with positive eponents and without fractional eponents: / ( ) ( ) / ( ) ( ) Solution: When factoring, always factor out the lowest eponent for each term. means and Eample: y 6 Solution: The lowest eponent for is - so y ( ) can be factored from each term. Leaving. Notice that for the eponent for the 6 term we take - (-) and get. For the term we take --(-) and get as our new eponent. When dividing two terms with the same base, we subtract the eponents (numerator eponent- denominator eponent). If the difference is negative then the term goes in the denominator. If the difference is positive then the term goes in the numerator. Eample: Simplify () 8 Solution: First you must distribute the eponent. 8. Then since we have two terms with as 8 the base we can subtract the eponents. Since -8 results in -5 we know that we will have in the denominator
10 Eample: Simplify ( ) Solution: First we must factor both the numerator and denominator. ( ) ( )( ). Then we can see that we have the term (-) in both the numerator and denominator. Subtracting eponents we get -= so the term will go in the numerator with as it s eponent. ( ) ( ) ( ) f. Eample: Factor and simplify 4( ) / ( ) / Solution: The common terms are and (-). The lowest eponent for is. The lowest eponent for (- ) is -/. So factor out / ( ) and obtain ( )./ [4( ) ] to ( ) / [4 ]. Leaving a final solution of (5 ).. This will simplify Problem Set VI Write without fractional eponents.. y /. f() = (6 ) 4 Write with positive eponents:. 4. y (4 ) ( ) 5. Factor then simplify: 6. 4 ( ) 8 / 7. 5 ( ) / ( ) 8. 6( ) 4( ) Simplify: (4 ) ( )( ) y 4 ( ) ( ) y 5 0 5
11 You will need to use your own paper to work the problems in this packet. You will turn in ALL of your work and the attached completed answer sheet. Answers only will result in no credit. AP CALC Summer Packet Name Problem Set I ) -int y-int ) -int y-int ) -int y-int 4) -int y-int 5) -int y-int Problem Set II ) y = ) y = ) y = 4) y = 5) y = 6) m = b = 7-9) (put on the same graph) 0) parallel, y = and perpendicular, y = Problem Set III ) f(g())= _ ) f( + 5) = ) g( + h) g() = 4) domain range 5) domain range 6) f()=_ g()= 7) f() = g() =
12 Problem Set IV ) holes vertical asymptotes -int horizontal asymptote ) holes vertical asymptotes -int horizontal asymptote ) holes vertical asymptotes -int horizontal asymptote 4) holes vertical asymptotes -int horizontal asymptote Problem Set V ) = _ ) = _ ) = _ Problem Set VI ) y = ) f () = ) f () = 4) y = 5) f () = 6) f () = 7) f () = 8) f () = 9) f () = 0) y = ) y =
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