AP Calculus AB Summer Assignment

AP Calculus AB Summer Assignment Name: When you come back to school, it is my epectation that you will have this packet completed. You will be way behind at the beginning of the year if you haven t attempted as much as you can. Shortly into the school year, we will test over this material. All of the material will have been material you ve seen before, but possibly just adapted in different ways. The summer assignment is broken down into seven tutorials. Each tutorial contains some definitions and some eample problems. You should read through all of these, and consult these eamples when you have questions. The end of the tutorial will have a section entitled homework. It is these seven homework sections that will be due at the beginning of the school year. The seven tutorials are: Tutorial #1 Linear Functions Tutorial # Function Characteristics Tutorial # Piecewise Functions Tutorial #4 Eponential and Logarithmic Functions Tutorial #5 Trigonometric Epressions and Equations Tutorial #6 Solving Inequalities with Sign Charts Tutorial #7 Comple Algebraic Simplifications All homework assignments should be completed on separate paper. Any graphs that you are asked to draw need to be drawn carefully. Round any approimate answers to three decimal places. If you run into any problems on this assignment over the summer, feel free to e-mail me over the summer. My e-mail address is tjkennedy@bluevalleyk1.org I look forward to seeing you in the fall! ~Mr. Kennedy 1

Tutorial #1 Linear Functions: Eample 1: Consider the line with slope - that passes through the point (4, -). a. Write the equation of the line in point-slope form. Point-slope form is y m 1 y, 1 where m is the slope and 1, 1 For this particular problem then, the equation is y 4. b. Graph the line. y is a point on the line. When you graph this line, first plot the point (4, -) and then use the slope of -. Here s what the final graph looks like in a standard viewing window: c. Find the area between this line and the -ais from when = - to =. The region in question is a trapezoid as shown in the diagram above. The area of a trapezoid is the average of the parallel sides times the perpendicular side. Since the slant height of the trapezoid runs from (-, 9) to (, 1) [We can get these points by subbing - and in for in the line s equation], the 9 1 10 trapezoid has parallel heights of 9 and 1, and a base of 4. Thus, its area is (4) (4) 0.

d. Find the equation of the line perpendicular to this line that passes through the point (-5, ). Leave your answer in point-slope form. Since the original line has slope, the perpendicular line has slope 1. 1 y 5. Thus, the answer is e. Find the equation of the line parallel to the line that passes through the point (1, 8). Leave your answer in point-slope form. Since the original line has slope, the parallel line has slope. Thus, the answer is y 1 8. Homework problems from tutorial #1: 1) Consider the line y. a. Graph the line. b. Find the area between the line and the -ais from when = 1 to = 4. ) Consider the line y 5. a. What is the slope of this line? b. Graph the line. c. Find the area between this line and the -ais from when = - to =. ) Consider the line 7. a. What is the slope of this line? b. Graph the line. 4) Consider the point (-, -9). a. Find the equation of the horizontal line that passes through this point. b. Find the equation of the vertical line that passes through this point. 5) Consider the line with slope 4 that passes through the point (-, 5). a. Find the equation of this line in point-slope form. b. Find the equation of a line perpendicular to this line that passes through the point (1, 9). Leave your answer in point-slope form. 4

Tutorial # Function Characteristics: Eample 1: Let f ( ) 5 7. (a) Use your calculator to help you draw a careful graph of the function. The calculator s graph is included above. Your graph should label the intercepts and etrema. You ll use your calculator s zero or ma or min tool to find these values. In this case, the -intercepts are (-.866, 0), (.11, 0), and (.655, 0). [Note: since eact values are not available, we will round to three decimal places. In AP Calculus, we will always round to three decimal places.]. The y-intercept is at (0, 7). The graph s maimum point occurs at (., 7.481) and the graph s minimum point occurs at (, -). (b) Give the domain of the function. The domain of this function is all real numbers,. (c) Give the equations of any asymptotes. Polynomial functions don t have any asymptotes. 5

(d) Determine when the function is increasing and decreasing. When describing where a function is increasing and where a function is decreasing, we use - coordinates only. Additionally, the AP Calculus curriculum epects you to include endpoints on increasing or decreasing intervals, if these endpoints are in the domain of the function. In this case, then, we say f is increasing,0.,. We say that f is decreasing over the interval 0.,. (e) Determine where the function has any maimum or minimum values. The maimum value of the function is 7.481 and it occurs when =.. The minimum value of the function is - and it occurs when =. (f) Determine when the function is positive and negative. The function is positive when the graph is above the -ais. This is.866,.11.655,. The function is negative when the graph is above the y-ais. This is,.866.11,.655. (g) Epress the end behavior of f using limit notation. (Use the statements lim f ( ) & lim f ( ). ) As the -values get very positive, the curve goes up, so lim f( ). As the -values get very negative, the curve goes down, so lim f( ). Eample : Let f( ). 1 (a) Use your calculator to help you draw a careful graph of the function. The calculator s graph is included above. Your graph should include the vertical asymptote at = 1, the y-intercept at (0, -) and clear labeling of the points where maimum and minimum values occur. These points are (-1, -) and (, 6). 6

(b) Give the domain of the function. The domain of this function is all real numbers ecept where the vertical asymptote occurs. We would,1 1,. write this as (c) Give the equations of any asymptotes. There is a vertical asymptote at = 1. There is no horizontal asymptote as the degree of the numerator is greater than the degree of the denominator, however there is a slant (oblique) asymptote. To find this asymptote, we divide by 1 and take the quotient. While we could use long division for this take, synthetic division is easier. As seen from the synthetic division below, the slant asymptote is at y = + 1. 1 1 0 1 1 1 1 4 (d) Determine when the function is increasing and decreasing. In this problem, we say the function is increasing over the intervals, 1,. The function is decreasing over the intervals 1,1 1,. Note that we do not bracket the = 1 as = 1 is not in the domain of this function. (e) Determine where the function has any maimum or minimum values. The function has a minimum value of 6, which occurs when =. The function has a maimum value of -, which occurs when = -1. While this may seem weird, it s important to remember that these points are merely local etreme values, and not absolute etrema. (We ll get into this idea more in chapter 4). Additionally, this idea of having a minimum with a larger value than a maimum is possible when the function isn t continuous for all real numbers (as is the case here). (f) Determine when the function is positive and negative. The function is positive when the graph is above the -ais. This is 1,. The function is negative when the graph is below the -ais. This is from, 1. (g) Epress the end behavior of f using limit notation. (Use the statements lim f ( ) & lim f ( ). ) As -values get larger and larger, the rational function continues to go up, so we say that lim f( ). As -values get increasingly negative (as we go to the left), the rational function is on its way down. So, we say that lim f( ). 7

Homework problems from tutorial #: For questions 1-5, complete the following steps: (a) Use your calculator to help you draw a careful graph of the function. (b) Give the domain of the function. (c) Give the equations of any asymptotes. (d) Determine when the function is increasing and decreasing. (e) Determine where the function has any maimum or minimum values. (f) Determine when the function is positive and negative. (g) Epress the end behavior of f using limit notation. (Use the statements lim f ( ) & lim f ( ). ) 1. f ( ).. g( ) e. h ( ) 7 4. 5. f ( ) 5 j( ) 1 Tutorial # Part These are all non-calculator questions. You would be epected to answer questions like these without your calculator. Recall that the formula for a circle centered at (0,0) is y r. If we wanted to graph this function on our calculators, we d have to solve for y. Solving this equation for y means we need to subtract the square root. (We can t forget the plus or minus when we take the square root!) This gives us The equation and then take y r. y r represents the top-half of a circle (or a positive semi-circle) and the equation y r represents the bottom-half of a circle (or a negative semi-circle). 6. Consider the equation y 6. a. What is the radius of this semi-circle? b. Graph the semi-circle. c. Find the area between the semi-circle and the -ais. 7. Consider the equation y 16. a. What is the radius of this semi-circle? b. Graph the semi-circle. c. Where is the function increasing and decreasing? d. What are the maimum and minimum values of this function? 8. Consider the equation y 5 ( 1). a. This equation represents a semi-circle, but it s not centered at (0, 0). What is the center of this semicircle? b. Draw a graph of this semi-circle. c. Find the area between this graph and the -ais between = 1 and = 6. 8

Problems from Tutorial # continue on the net page: In Calculus, we will frequently use tables to learn more information about functions. In the table you are given below, you are given si selected -values, and corresponding values of f() and g(). For eample, if you were to look at the column where = 0, you would see that f() = - and g() =. This means that the point (0, -) is on the graph of f() and the point (0, ) is on the graph of g(). It also means that f(0) = - and g(0) =. Use this information to help answer the questions below. Suppose f() and g() are continuous functions with domains of [0, 5]. 0 1 4 5 f( ) - 4 6 8 11 g ( ) 5-6 1 0-4 9. Compute f(1) g(1). 10. Compute f (1) g() f () g(1). 11. Compute f( g (0)). 1. Compute g( f (1)). 1. Compute f( g (4)). 14. Compute g( f(1)). Use the table to answer questions 9 18. 15. Write the equation of the line, in point-slope form, with slope 7 at the point corresponding to g(). 16. Write the equation of the line, in point-slope form, perpendicular to the line in #15, and passing through the same point. 17. Compute g(1) f(1), and eplain why this answer is different from your answer to #1. 18. Suppose f ( ). (a) Evaluate f() f( 1). (b) Evaluate f() f(0). 0 9

Tutorial # Piecewise Functions In Pre-Cal, you looked at the concept of graphing specific parts of functions. This particular tool is very helpful to us in the study of calculus. Our focus here will be on graphing piecewise functions, and evaluating them at specific points. Eample 1: Graph the function:, if 1 f( ) sin( ), if 1 In Pre-Calculus, the way that we graphed piecewise functions is that we graphed both of the original functions and then kept the part of the domain that was requested in the function. Your procedure for solving this problem would be to graph, and only keep the part of the line to the right of = 1 (i.e. where > 1), and then to graph the sinusoid and only keep the part of the sinusoid to the left of = 1. In Calculus, there are a few things that you ll be interested with regards to the piecewise function. First, you ll need to be able to describe the behavior around the function s discontinuity ( = 1). Second, you ll need to be able to describe the type of discontinuity (jump). A jump discontinuity has a gap. See the piecewise graphed above at = 1. A removable discontinuity looks like a point has been picked out of a function and either tossed off of the graph or placed somewhere else on the same vertical line where the removal occurred. Pictures of removable discontinuities: 10

, if 0 Eample : Consider f( )., if 0 (a) Find f, f (0), f (1) and f (). 4 To compute f, we are being asked for the y-value of the function when = -/4, so it s 4 important to determine which piece of the piecewise function we re supposed to use here. In this case, since / 4 is less than 0, we are using the piece. So, f 9. 4 4 When = 0, the sin domain of the other piece does not. sin (0) sin(0) 0. When = 1, sin (1) sin 0. piece is being used, as the domain of this piece includes zero, while the When =, the function is not defined, as = does not fit in the domain of either piece of this function. (b) Find all values of for which f( ) 0. This is a significantly harder question than (a), because this question is asking for what values of does the function have y-values of 0? Since either piece of the function could theoretically have y- values of 0 (remember, the piecewise function restricts what -values go with which functions, not which y-values go with which functions), we have to set each piece equal to 0 and solve both. Set 0. This leads to. So, the linear piece has a y-value of 0 when the - value is. However, the domain of this piece is 0, so = is not in our domain, and is therefore, not a solution to our equation. Set 0. We can solve this by factoring. Factor this into ( + )( 1) = 0. This gives us solutions of = - and = 1; however, the domain of this function is only 0, so only = 1 fits our domain. The only solution to the entire equation is = 1. 11

Homework problems from tutorial #: if 0 1. Consider the piecewise function, f ( ) 5 if 0. if 0 (a) Draw a graph of f(). (b) Does the function have a jump or a removable discontinuity at = 0? Eplain. (c) Find f(0). (d) Find f(10). (e) Find the value(s) of for which f( ) 10. (f) Calculate lim f( ). (g) Calculate lim f( ). 1 if 1. Consider the piecewise function, f( ) e if 1 (a) Draw a graph of f(). (b) Does the function have a jump or a removable discontinuity at = 1? Eplain. (c) Find f(1). (d) Find f(-). (e) Find the value(s) of for which f( ). (f) Calculate lim f( ). (g) Calculate lim f( ).. Draw an eample of a jump discontinuity and removable discontinuity; then, eplain the difference between the two. 1

Tutorial #4 Eponential and Logarithmic Functions Concept 1 Eponential Functions In Calculus, we will deal a little bit with eponential functions. An eponential function is defined by the equation, f ( ) a. In this definition, a must be positive and not equal to 1, while must be a real number, not equal to zero. Eample 1: Let f( ) 16. Evaluate f (0), f (), f (1/ ), f ( 1),and f ( / 4). Solutions: 0 f (0) 16 1. (Anything to the zero power is 1.) f () 16 56. 1 f 1 16 16 4. (Remember that anything to the ½ power is the same as finding the square root.) f 1 1 1 16. 1 1 16 16 (Any negative eponent can be made positive by moving the base and the power to the other part of the fraction.) f 1 1 1 1 4 16 16 8 / 4 16 / 4 4 Eample : Let f( ), and (a) Find ( g( f (1)). g 1 ( ). Solution: Since you want to calculate g(f(1)), you need to find f(1) first. Once you find f(1), substitute 1 1 that result into g. f(1). g() 8. g (b) Find ( ). f 1 g 1 1 Solution: ( ). f bases, you may subtract the eponents. Since you are dividing two eponential epressions with like 1

Eample : Solve the eponential equations: (a). Solution: This eponential equation can be solved easily by equating bases. Since both the left and right hand side of the equations can be epressed as powers of, rewrite as 5. Thus, you can rewrite the 5 equation. Thus, if you try and match both sides of the equation = 5. (b) 4 8 Solution: You can rewrite both sides of the equation as powers of. Rewrite 4 as. Then, you d have. Then =, so = /.. Rewrite 8 as (c) 8 5 1 16 Solution: Again, rewrite both sides as a power of. (5) 1 4, then 6 + 15 = -4, 4 19. 6 The main motivation for logarithmic functions stem from problems similar to what you saw in eample. In eample, we were able to solve all of our problems by simply equating bases. However, what happens in the situation where bases can t be equated? For eample, what happens if we want to solve 5 17? Clearly, we can t rewrite 17 as a power of 5. When mathematicians encountered this problem, they knew that getting the eponent out of the variable was a must. The way they removed the eponent from the variable was by taking either the logarithm or natural logarithm of both sides (it doesn t matter which one you use as long as you use it on both sides of the equation.) Logarithmic functions are the inverse functions of eponential functions. In other words, if y a, then log ( y a ). Eample 4: Rewrite the following eponential equations in logarithmic form. a) b) 5 1 c a b 5 Solution: 1 1 5 5 5 log5. Solution: b a c log b c. a Properties of logarithms: b a ln( a ) bln a ln( a) ln( b) ln( ab) ln( a) ln( b) ln b ln( e) 1 ln(1) 0 f ( ) ln b log b ln( e ) f ( ) loga b ln a log a 14

Eample 5: Rewrite the following epressions using properties of logarithms. e a) ln e 1 b) e e e ln ln ln 1. e e 1 1 Solution: e ln ln( e) Solution: Eample 6: Solve the following equations: a) 5 7 1 1 1 1 ln ln( ) ln( ) ln( ) e e e e Solution: Start by taking the logarithm or natural log of both sides. Then, we will have ln(5 ) ln(7). Now, use a property of logarithms to pull the eponent out front of the lefthand epression. So, then we ll have ( ) ln 5 ln 7. Then, solve for, and you ll get ln 7 ln 7 ln 7 ln 5. ln 5 ln 5 b) ln 4 ln(5) 4 Solution: First use properties of logs to rewrite ln( 4) ln(5) as ln. Then, we 5 4 want to rewrite to eliminate the natural log. Thus, e. If we want to solve this for, 5 we d need to start by multiplying through by the denominator Eample 7: Solve for y: ln y ln. 15 5e 4 5e 4. Solution: This equation is tricky to solve, but we are first going to need to get rid of the negative on the left-hand side of the equation. So, let s divide both sides through by -1. This gives us ln y ln. Then, we need to undo the natural log on the left hand side of the equation. This means we now have the statement y e ln. Net, we are going to make a funny simplification. Since we have added eponents in the right hand side of the equation, we can ln rewrite that side to have multiplied bases. In other words, e e e ln. This is useful because ln ln we know that e. So, e e. This means our overall equation is y e. Net, we need to undo the absolute value bars. We do this by recognizing that the right hand side of the equation may be either positive or negative, so y e. Isolating y gives us y e.

Homework from Tutorial #4: 1. Eplain why f ( ) e is an eponential function while f ( ) is not an eponential function.. Eplain why f ( ) a is not an eponential function in the situations where a = 1 or where = 0. What kind of function would f() be in these situations?. Let f( ) 8. Evaluate f (0), f (), f ( ),and f ( 5/ ). 1 4. Evaluate log. 7 5. Simplify the following epressions using properties of logarithms: 8 11 a (a) ln( e ) ln( e ) (b) log ( ) (c) ln( e) ln(1) (d) 1 ln y e 6. Solve the following equations for y: (a) y ln y ln (b) e e 5 4 e 7. Find f(1) f(0) if f( ). 4 1 8. Evaluate when = 0: 4 e (1 e ) 16

Tutorial #5 Trigonometric Epressions and Equations Eample 1: Use the unit circle to evaluate the following epressions: (a) (b) (c) (d) 5 1 cos. 7 cos 7 6 cot. 6 7 1 1 sin 6 19 1 7 5 1 csc csc csc csc csc / sin(0) 0 tan(1 ) tan(10 ) tan(8 ) tan(6 ) tan(4 ) tan( ) tan(0) 0. cos(0) 1 Eample : Evaluate the following inverse trigonometric functions. (Remember that inverse trigonometric functions use ratio as inputs, and output angles.) 1 (a) sin. (Keep in mind that the range of arcsine is from,. ) The angle in the range that has a sine of / is 60 or /. 1 (b) tan. (Keep in mind that the range of arctangent is from,. ) The angle in the range that has a tangent of is the angle that has a sine of / and a cosine of 1/. (c) cos 1 1 (Keep in mind that the range of arccosine is from 0,. ) The angle in this range that has a cosine of -1/ is /. 17

(d) sin 1 4 (Keep in mind that the range of arcsine is from,. ) The angle in this range that has a sine of / is / 4. Many of you will want to use either 5 7 or as the answer here. However, both of 4 4 these are wrong, because they don t fit into the range. Homework from Tutorial #5: 1. Evaluate the following epressions without the use of a calculator. (a) 7 cos 6 (b) cot (c) csc 4 (d) tan( ) (e) cos 1 (f) 1 tan 1 (g) sin 1 1 (h) cos 1. Eplain the difference between y csc and y 1 sin.. Evaluate the following epressions, if f ( ) sin( ). (a) Find f. (b) Find f. 4 (c) Find f. 18 18

Tutorial #6 Solving Inequalities with Sign Charts: There will be multiple times in AP Calculus when we develop an epression and must determine when that epression is positive or negative. In most cases, we ll have to do this without a calculator. Basically, a function can switch from positive to negative or negative to positive at two different times. The function can either cross the -ais (so we consider points where the function equals zero) to change signs or the function can have a vertical asymptote. Consider the following eamples: Eample 1: Solve 0. Solution: This function is undefined when = 0, so this will automatically make our number line as a possible location where this switches sign. To determine any other values for our number line (we ll call them critical points), set 0 and solve. You can either get common denominators or move one term to the other side to solve this equation. I think it s easiest to move / over to the other side. So, we have. Then, multiply 4 both sides by. So, we have. Then, solving gives us 4. (We get the plus or minus because we are undoing an even power). Net, we make our number line. X X X 4 4 0 Then, we choose number between each interval and determine whether the result is positive or negative. 4 So, for the interval,, choose a number like 5. If we put -5 back into the original equation, we get 5 15, which is a negative number. Thus, we fill out the first part of the number line with 5 5 negative signs like this: X X X 4 4 0 For the interval 4,0, we can choose -1. This gives us 1 1 0. ( 1) 4 For the interval 0,, we can choose 1. This gives us 1 1 0. 1 For the interval 4,, we can choose. This gives us 81 0. Then, our number line looks like: X X X 4 4 0 4 4 So, our solution is, 0,. 19

Eample : Solve e 6e 6e 0. Solution: Start by combining like terms. This gives us e 1e 0. Then, factor out the e term. This e 1 0. This epression is always defined, so we only need to consider values that make gives us either term equal to 0 as critical points. Start by setting e 0. This leads to no values of as a positive base raised to a power is always positive. (Alternatively, you could try to solve this taking the natural logs of both sides, but you would get = ln(0), and you can t take the natural log of a number that s not positive.) Then, set 1 0. This has a solution of = -6. So, this is our only critical point. X 6 Sub in a value in, 6 take = -10. Sub in a value in 6, take = 0. 10 10 10 10 10 10 8 ( 10) e 6e 6e 0e 1e 8e 0. 10 0 0 0 0 0 0 (0) e 6e 6e 0e 1e 1e 1 0. e The solution is, 6. X 6 Eample : Solve 0. Solution: The zeros of this inequality occur when = - and when =. Make a number line. X X Sub in a value in, 5 ( 5 ) 5 ( 8) (64) 0. Sub in a value in, 0 (0 ) 0 ( ) (9) 0., 5 (5 )(5 ) 7() 0. Sub in a value in The solution is,,. [Note: we can t write the solution as, =.] as the function isn t positive at Eample 4: Solve sin 0 in the interval 0,. Solution: Solve sin( ) 0. The angles with a sine of zero are 0 and. We need to find the -values that correspond 0 and. If we set 0&, we get = 0 and. (continues onto net page) 0

X 0 X In 0, use sin sin 1 0. 4 4 In, use sin sin 1 0 4 4 X 0 X So, the answer is,. 1 Eample 5: Solve cos 0 in the interval [0,16). 8 1 Solution: Solve cos. Now, use the unit circle to help you. Think of the angles whose cosines are 8 1. These angles are & 4. 4 1 So, we have cos cos cos. If we set 8 4 &, we can get the values of that solve this inequality. So, we have, we can 8 8 8 16 16 4 multiply both sides by 8 to get. Additionally,. 8 In In In X X 16 16 1 4 1 1 1 0, 4 cos (4) cos cos 0 0. 8 8 16 1 1 1 1,, pick = 8. Then, cos (8) cos 1 0. 8,16, pick = 1. Then, 1 1 1 1 1 cos (1) cos cos 0 0. 8 8 So, our number line looks like: X X 16 16 So, our answer is,. Homework problems from tutorial #6: 1

Solve the following inequalities: 1) 100 50 ) ( ) 0 ) 1 5 0 4) e 1 0 cos 0 in the interval 5) 0, 6) sin 0 6 in the interval [0, 1) cos 0 in the interval [0, ) 7)

Tutorial #7 Comple Algebraic Simplifications: f ( ) g( ) h( ) j( ) Eample 1: Determine what f ( ) is if f g e h e j ( ) 1, ( ), ( ), ( ). Solution: By substitution, we immediately have After distributing, we have f g h j e e f ( ) ( 1) ( ) ( ) ( ) ( ) ( 1)( ) ( )( ). ( 1) e e e e e Combining like terms and factoring the numerator gives us. e e ( 1) e. ( 1) Eample : Solve: e e 0. Solution: Start by factoring the common factor out of each term. Some of you will notice that the common factor here is e. However, the process is doable even if you only recognize e as the common factor. Factoring out e 0. Setting each term equal to zero, makes 0, which we e gives us can solve by factoring out an, yielding, ( ) 0. So, we have solutions at 0,,. Setting e 0 produces an interesting case, as anything raised to a power won t ever equal zero!! So our only solutions are 0,,. Comment: You will make very heavy use of eponent properties in this section. A review of these properties are listed here. n 1 Negative Eponent Property: a n. a Raising a quantity to a negative eponent is equivalent to writing one over the quantity with a positive eponent. m / n n m n Rational Eponents Property: a a a m Raising a quantity to a rational eponent is the same as raising the quantity to the power of the rational eponent s numerator and then taking the root of that quantity that corresponds to the number of the denominator. Eample : Rewrite 7 5r ( r 1) so that no variables are in the denominator and no radicals are in your answer. / 7 5r Solution: First rewrite the denominator as ( r 1). So, we have. Then to move that quantity / 7 ( r 1) out of the denominator, we need to rewrite it with a negative eponent, and this is our final answer: 5r 5rr1 / 7 / 7 ( r 1)

Homework from Tutorial #7: f ( ) g( ) h( ) j( ) 1) Determine what f ( ) assistance.] is. Simplify your answer as much as possible. [Consult e 1 for a) Let b) Let f h g j ( ) 5, ( ), ( ), ( ). f h j e g e ( ), ( ), ( ), ( ). ) Solve the equation employing techniques you know (factoring, common denominators): a) e e 0 b) 4 5 5 5 e 4 e = 0 c) 1 0 ) Rewrite the radical epressions using eponents. Your answers will contain neither radicals nor fractions. Your answers, however, may contain negative eponents. (In fact, some of them should ) a) 5 b) 9 5 7 c) 1 d) 9 1 e) 7 4) Rewrite each radical as something to a rational eponent (rational eponents are just eponents with fractions), then use properties of eponents to combine like terms: a) y y b) z z z c) 5) Rewrite the following epressions so that no variables are in the denominator. 5 a) 7 1 b) 4 8 c) ( ) d) e 6) Solve for y: (a) y (b) 8 5 1/ y 5 1/ / 7) Calculate the value of C (without using a calculator) if () 8 C 4