Algebra. Exponents. Absolute Value. Simplify each of the following as much as possible. 2x y x + y y. xxx 3. x x x xx x. 1. Evaluate 5 and 123

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1 Algebra Eponents Simplify each of the following as much as possible y + y y y + y Absolute Value Evaluate 5 and 1. Eliminate the absolute value bars from 8. List as many of the properties of absolute value as you can. 006 Paul Dawkins 1

2 Radicals Evaluate the following Convert each of the following to eponential form Simplify each of the following y Assume that 0 and y 0 for this problem y Rationalizing Rationalize each of the following. y + y 006 Paul Dawkins

3 t +. t 4 Functions Given f ( ) = and g( ) = 4 find each of the following. (a) f ( ) (b) g ( ) (c) f ( ) (d) g ( 10) (e) f ( t ) (f) f ( t ) (g) f ( ) (h) f ( 4 1). Given f ( ) = 10 find each of the following. (a) f ( 7) (b) f ( 0) (c) f ( 14). Given f ( ) = + 10 and g( ) 1 0 f g f g (a) ( f g)( ) (b) ( ) = find each of the following. (c) ( fg)( ) (d) ( f g)( 5) (e) ( )( ) (f) ( g f )( ) Multiplying Polynomials Multiply each of the following. ( 7 4)( 7+ 4). ( 5). ( + ) + 4. ( ) 5. ( + )( 9+ 1) 006 Paul Dawkins

4 Factoring Factor each of the following as much as possible Simplifying Rational Epressions Simplify each of the following rational epressions Graphing and Common Graphs Sketch the graph of each of the following. y = + 5. y ( ) = + 1. y = ( ) f = Paul Dawkins 4

5 5. y y = ( y ) = 4 7. y y = 0 8. ( ) ( y ) = 1 9. ( + 1 ) ( y ) 10. y = = y = y = Solving Equations, Part I Solve each of the following equations... = = = 0 Solving Equations, Part II Solve each of the following equations for y. y 5 = 6 7y. ( ) 5y + sin = y y = Paul Dawkins 5

6 Solving Systems of Equations Solve the following system of equations. Interpret the solution. y z = + y+ z = 5 4y+ z = 10. Determine where the following two curves intersect. + y = 1 y =. Graph the following two curves and determine where they intersect. = y 4y 8 = 5y+ 8 1 Solving Inequalities Solve each of the following inequalities > > < Absolute Value Equations and Inequalities Solve each of the following. + 8 = 006 Paul Dawkins 6

7 . 4 = = < > > 4 Trigonometry Trig Function Evaluation One of the problems with most trig classes is that they tend to concentrate on right triangle trig and do everything in terms of degrees. Then you get to a calculus course where almost everything is done in radians and the unit circle is a very useful tool. So first off let s look at the following table to relate degrees and radians. Degree π π π π π Radians 0 π π 6 4 Know this table! There are, of course, many other angles in radians that we ll see during this class, but most will relate back to these few angles. So, if you can deal with these angles you will be able to deal with most of the others. Be forewarned, everything in most calculus classes will be done in radians! Now, let s look at the unit circle. Below is the unit circle with just the first quadrant filled in. The way the unit circle works is to draw a line from the center of the circle 006 Paul Dawkins 7

8 outwards corresponding to a given angle. Then look at the coordinates of the point where the line and the circle intersect. The first coordinate is the cosine of that angle and the second coordinate is the sine of that angle. There are a couple of basic angles that are π π π π π commonly used. These are 0,,,,, π,, and π and are shown below along 6 4 with the coordinates of the intersections. So, from the unit circle below we can see that π π 1 cos = and sin =. 6 6 Remember how the signs of angles work. If you rotate in a counter clockwise direction the angle is positive and if you rotate in a clockwise direction the angle is negative. Recall as well that one complete revolution is π, so the positive -ais can correspond to either an angle of 0 or π (or 4π, or 6π, or π, or 4π, etc. depending on the direction of rotation). Likewise, the angle 6 π (to pick an angle completely at random) can also be any of the following angles: π 1π π + π = (start at then rotate once around counter clockwise) π 5π π + 4π = (start at then rotate around twice counter clockwise) π 11π π π = (start at then rotate once around clockwise) Paul Dawkins 8

9 π π π 4π = (start at then rotate around twice clockwise) etc. π π In fact can be any of the following angles + π n, n= 0, ± 1, ±, ±, In this case 6 6 n is the number of complete revolutions you make around the unit circle starting at 6 π. Positive values of n correspond to counter clockwise rotations and negative values of n correspond to clockwise rotations. So, why did I only put in the first quadrant? The answer is simple. If you know the first quadrant then you can get all the other quadrants from the first. You ll see this in the following eamples. Find the eact value of each of the following. In other words, don t use a calculator π sin and π sin 7π cos 6 and 7π cos 6 π tan 4 and 7π tan π sin 4 5π sec 6 4π tan Trig Evaluation Final Thoughts As we saw in the previous eamples if you know the first quadrant of the unit circle you can find the value of ANY trig function (not just sine and cosine) for ANY angle that can be related back to one of those shown in the first quadrant. This is a nice idea to remember as it means that you only need to memorize the first quadrant and how to get the angles in the remaining three quadrants! 006 Paul Dawkins 9

10 In these problems I used only basic angles, but many of the ideas here can also be applied to angles other than these basic angles as we ll see in Solving Trig Equations. Graphs of Trig Functions There is not a whole lot to this section. It is here just to remind you of the graphs of the si trig functions as well as a couple of nice properties about trig functions. Before jumping into the problems remember we saw in the Trig Function Evaluation section that trig functions are eamples of periodic functions. This means that all we really need to do is graph the function for one periods length of values then repeat the graph. Graph the following function. y = cos( ). y = cos( ). y = 5cos( ) 4. y = sin( ) 5. y = sin 6. y = tan ( ) 7. y = sec( ) 8. y = csc( ) 9. y = cot ( ) 006 Paul Dawkins 10

11 Trig Formulas This is not a complete list of trig formulas. This is just a list of formulas that I ve found to be the most useful in a Calculus class. For a complete listing of trig formulas you can download my Trig Cheat Sheet. Complete the following formulas. sin ( θ) cos ( θ) tan θ + 1=. ( ). sin( t ) = + = 4. cos( ) = cos ( ) = sin ( ) = (Three possible formulas) (In terms of cosine to the first power) (In terms of cosine to the first power) Solving Trig Equations Solve the following trig equations. For those without intervals listed find ALL possible solutions. For those with intervals listed find only the solutions that fall in those intervals. cos( t ) =. cos( ) t = on [ π, π]. sin( 5 ) = 4. sin( 5 + 4) = 5. sin( ) = 1 on [ ππ, ] 6. sin( 4t ) = 1 on [0, π ] 006 Paul Dawkins 11

12 7. cos( ) = 8. sin( ) = cos( ) 9. sin( θ) cos( θ ) = ( w) ( w) ( w) sin cos + cos = 0 1 ( ) ( ) cos + 5cos = 0 5sin( ) = cos = sin( ) = 7 Inverse Trig Functions One of the more common notations for inverse trig functions can be very confusing. First, regardless of how you are used to dealing with eponentiation we tend to denote an inverse trig function with an eponent of -1. In other words, the inverse cosine is cos. It is important here to note that in this case the -1 is NOT an denoted as ( ) eponent and so, cos ( ). cos 1 1 In inverse trig functions the -1 looks like an eponent but it isn t, it is simply a notation that we use to denote the fact that we re dealing with an inverse trig function. It is a notation that we use in this case to denote inverse trig functions. If I had really wanted eponentiation to denote 1 over cosine I would use the following. ( cos( ) ) ( ) 1 = cos There s another notation for inverse trig functions that avoids this ambiguity. It is the following. cos = arccos sin tan ( ) ( ) ( ) ( ) = arcsin ( ) ( ) = arctan ( ) 006 Paul Dawkins 1

13 So, be careful with the notation for inverse trig functions! There are, of course, similar inverse functions for the remaining three trig functions, but these are the main three that you ll see in a calculus class so I m going to concentrate on them. To evaluate inverse trig functions remember that the following statements are equivalent. θ = cos = cos θ θ θ ( ) ( ) ( ) sin ( θ) ( ) tan ( θ) = = sin = = tan In other words, when we evaluate an inverse trig function we are asking what angle, θ, did we plug into the trig function (regular, not inverse!) to get. So, let s do some problems to see how these work. Evaluate each of the following... cos cos 1 1 sin 4. tan ( 1) cos cos 1 π sin sin 4 ( ) 7. tan tan ( 4). 006 Paul Dawkins 1

14 Eponentials / Logarithms Basic Eponential Functions First, let s recall that for b > 0 and b 1 an eponential function is any function that is in the form f = b ( ) We require b 1 to avoid the following situation, f = = ( ) 1 1 So, if we allowed b 1 we would just get the constant function, We require b > 0 to avoid the following situation, 1 f ( ) = ( 4) f ( 4) 1 = = 4 By requiring b > 0 we don t have to worry about the possibility of square roots of negative numbers. Evaluate f ( ) = 4, g( ) 1 = 4. Sketch the graph of f ( ) = 4, g( ) and h( ) 4 1 = 4. List as some basic properties for f ( ) = b. 4. Evaluate f ( ) = e, g( ) = e and h( ) 5. Sketch the graph of f ( ) = e and g( ) = at =, 1,0,1,. and h( ) 4 = on the same ais system. 1 = 5e at, 1,0,1, = e. =. Basic Logarithmic Functions Without a calculator give the eact value of each of the following logarithms. (a) log16 (b) log4 16 (c) log (d) log9 (e) log1 6 (f) log Without a calculator give the eact value of each of the following logarithms. (a) ln e (b) log1000 (c) log Paul Dawkins 14

15 7 (d) log1 (e) log Logarithm Properties Complete the following formulas. log b b =. log 1 b =. log b = b 4. log b b = 5. log b y = 6. log b = y r 7. log ( ) b = 8. Write down the change of base formula for logarithms. 9. What is the domain of a logarithm? 10. Sketch the graph of f ( ) = ln ( ) and g( ) log( ) =. Simplifying Logarithms Simplify each of the following logarithms. 4 5 ln yz. 4 9 log y 006 Paul Dawkins 15

16 . log + y ( y) Solving Eponential Equations In each of the equations in this section the problem is there is a variable in the eponent. In order to solve these we will need to get the variable out of the eponent. This means using Property and/or 7 from the Logarithm Properties section above. In most cases it will be easier to use Property if possible. So, pick an appropriate logarithm and take the log of both sides, then use Property (or Property 7) where appropriate to simplify. Note that often some simplification will need to be done before taking the logs. Solve each of the following equations. 4 e = 9 t t. 10 = e z = e = 0 Solving Logarithm Equations Solving logarithm equations are similar to eponential equations. First, we isolate the logarithm on one side by itself with a coefficient of one. Then we use Property 4 from the Logarithm Properties section with an appropriate choice of b. In choosing the appropriate b, we need to remember that the b MUST match the base on the logarithm! Solve each of the following equations. ( ) 4log 1 5 =. + ln + = Paul Dawkins 16

17 . ( ) ( ) ln ln 1 = 4. ( ) log + log = Paul Dawkins 17

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