Lesson A  Natural Exponential Function and Natural Logarithm Functions


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1 A Lesson A  Natural Exponential Function and Natural Logarithm Functions Natural Exponential Function In Lesson 2, we explored the world of logarithms in base 0. The natural logarithm has a base of e. e is approximately The number e was discovered by a great 8th century mathematician named Euler. This famous irrational number is useful for determining rates of growth and decay. The derivation of e is beyond the scope of this course, but the expresson + n closer and closer to e, as the value substituted for n gets larger and larger. n produces results The natural exponential function is f(x) ex. Your scientific calculator should have a key for finding ex. You can also use the approximate value of e for computations. Below is the graph for f(x) ex f(x) ex x ex Rules of exponents apply to the exponential function. Study the examples of factoring. Example e2x  ex ex ex (ex  ) ex ex  Sometimes it is easier to factor by making a substitution. Let y ex. Example 2 4e2x  3ex 2ex Substituting yields 4y23y 2y y(4y  3) 2y 4y y  3 Don t forget to substitute back again. 2 4ex Practice Problems Factor the following. ) e2x  2) e2x  ex  6 3) e2x  ex  2 4) 2e2x  5ex  3 Solutions ) e2x  2) e2x  ex  6 3) e2x  ex  2 4) 2e2x  5ex  3 y2  y2  y  6 y2  y  2 2y25y  3 (y  )(y + ) (ex )(ex + ) (y  3)(y + 2) (ex  3)(ex + 2) (y  2)(y + ) (ex  2)(ex + ) (2y + )(y  3) (2ex + )(ex  3)
2 A2 The natural exponential function is used in every area of science. Example 3 Suppose that the number of bacteria present in a culture is given by N(t) 000 e (.)t. How many bacteria would be in the culture when t 4 hours? 000 e(.)(4) 000 e(.4) 49.8 rounds to,492 bacteria Practice Problems ) How many bacteria would be in the culture above when t 0 hours? 2) How long will it be before the number of bacteria reaches 0,000? Try to solve this by plugging in values for t. We will learn how to solve this a different way soon. Solutions ) 000 e(.)(0) 000 e() 2,78.3 rounds to 2,78 bacteria 2) try t 5 hours 000 e(.)(5) 000 e(.5) rounds to 4482 bacteria try t 20 hours 000 e(.)(20) 000 e(2) 7389 bacteria try t 24 hours 000 e(.)(24) 000 e(2.4),023 bacteria Natural Logarithm Function It is the inverse of the exponential function, which is f(x) ex. You may have tried different values, but you should have determined that the time to reach 0,000 bacteria is somewhat less than 24 hours. The natural logarithm function is f(x) ln(x). In stands for natural log. There are several properties and laws of the natural log function which you need to memorize. Use the directions that came with your scientific calculator to find and use the natural log key. (It may be marked LN.) Check the following relationships with your calculator, choosing any value you like for the variables. ) ln 0 2) ln e 3) ln ex X 4) elnx x when x 0 5) ln xy ln x + ln y 6) ln x/y In x  ln y 7) ln xa a In x Example 4 Determine the value of ln 2e in simplest terms. ln 2e ln 2 + ln e Law #5 above.69 + Use your calculator to find the natural log of 2. You should know the natural log of e from #2 above. ln 2e.69 Add to find the solution.
3 A3 Example 6 Determine the value of 2 ln 33 ln 2 in simplest terms. 2 ln 33 ln 2 ln 32  ln 23 Law #7 above ln 9  ln 8 Simplify radicals ln 9/8 ln.25.8 Law #6 above Divide and find the ln. The answer is rounded. Example 7 Determine the value of ln 5 + ln 3  ln 5 in simplest terms. ln 5 + ln 3  ln 5 ln (5 3)  ln Law #5 above 5 ln 5  ln 5 Multiply 0 Subtract Practice Problems Determine the value of each expression in simplest terms. Round your answers to the nearest hundredth. ) In 4e2 2) In 8  In 2 + In 5 3) 4 In x + 2 In y 4) In 5 Solutions ) In 4 e2 ln 4 + ln e2 2) In 8  In 2 + In 5 ln 8/2 + ln ln e ln e () ) 4 In x + 2 In y ln x4 + ln y2 4) In 5 ln 52 ln x4 y2 ln x4y2 /2 ln 5 ln 4 + ln 5 ln 4 5 ln /2 (.609).80 Example 8 Recall that on page 222, you were asked to find out how long it would be before the number of bacteria reached 0,000. Let s work that problem a different way using the natural logarithm function. 000 e.t 0,000 e.t 0 In (e.t) In (0).t In 0 Taking the natural logarithm of both sides. t In 0 23 hours.
4 Exponential and logarithmic functions can be manipulated in algebraic equations. Remember that as long as we do the same thing to both sides of an equation, we do not change the value of the equation. In the examples below, find the natural log of each side in order to simplify exponents and put the equation in a form that is easier to manipulate. A4 Example 9 4 X2 9 ln 9 ln 4 X2 X 2 ln 4 ln 9 X 2 ln 9 ln X.26 Example 0 4x+2 72x In (4x+2 ) In (72x ) (X + 2) In 4 (2X  ) In 7 X In In 4 2X In 7  In 7 X In 42X In 72 In 4  In 7 X (In 42 In 7) 2 In 4  In 7 X 2 In 4  In 7 In 42 In X
5 B Lesson B  Limits Limits in life are boundary points. Usually they denote the highest or top number allowable. Speed its are the highest speed allowable which cars may tavel without breaking the law. A supermarket it of 2 quarts of strawberries on their store special means that you can buy up to 2 quarts per person and no more. In science an instrument might be accurate within /00th of an inch. In each case, the it is a real number. Mathematical its are similar, but are not identical, to the its we see in everyday life. In math, we refer to the it of a function. A function s it may or may not exist. If the it does exist, then the it will be a real number and it will be unique. Example Consider the following sequence:, /2, /3, /4, /5, /6,... /000000,... As we look at these numbers, what is happening to them? They are getting smaller and smaller, right? How small will they get? Will they reach 0? No, but they will get so close to zero as to be indistinguishable. Therefore, the mathematical it of these numbers is 0. In this case, the it exists and is uniquely 0. It is not approaching any other number except 0. Example 2 Now let s look at another sequence:, 2, 3, 4, 5, 6,... 0,000,... 00,000, What is happening to these numbers? They are getting larger and larger and are approaching infinity. Is there a it? NO! If you say that the it is 00,000,000,000,003, then I can always find a bigger number in the sequence. Therefore, the it does not exist. Example 3 In geometry, consider a circle which has a polygon inscribed inside. If you begin with a 3sided polygon (triangle), and continue to inscribe polygons with larger and larger numbers of sides, what happens? Notice that the area of the inscribed polygon gets closer and closer to the area of the circle. Does the inscribed polygon become a circle? No, but it gets close enough as to be indistinguishable. Therefore, the it exists and is uniquely the area of the circle. Historically, Archimedes used this exact reasoning to determine the area of circles before the discovery of pi. Now let s look at some functions on a graph, and see if we can understand its more fully. Consider the following function. f(x) Example 4 f(x) X +
6 B2 A it in a graphical sense represents the anticipated height of a function. What would be the it of f(x) as we allow our X values to get closer and closer to 0? We could approach 0 by using negative numbers or positive numbers. A chart below shows numbers which are approaching X 0 from both perspectives. X f(x) X f(x) looking at positive values for X looking at negative values for X Either way we look, we see that as the X value of this function approaches 0, the f(x) value or the height of the function will be approaching. Indeed, f(0)! Thereore, the it exists and it is uniquely. Example 5 f(x) X + for all X except x 0 Graphically, we now have: f(x) Now what is the it as X approaches 0? It is still. Even though the value at X 0 is undefined, we can get SO close to X 0, and it still appears that f(x) is SO close to. How close is close enough? That is a discussion for Caluculus. f(x) Let s try a few more: Example 6 f(x) X2 What is the it of f(x) as X approaches 2? This one is simple. The graph will show the exact value of 4 when x 2. You could look at values on both sides of 2, but your answer will be the same. The it exists and it is uniquely equal to 4.
7 B3 Example 7 X 2 X 2 X 2 The graph looks like: f(x) What is the it of f(x) as X approaches 2? We will need a table here, because the value of 2 cannot be substituted into the function. Therefore, the it of f(x) 3. Notice that the numerator of this function can be factored into (X + )(X  2). Then, you can cancel the common factors, and you get (X ). The graph looks like f(x) X +, except that there is a hole at X 2. In fact, if you substitute x 2 into the new equation, f(x) X +, you get f(2) Now, let s work with mathematical notation to accurately describe what we have done so far. The appropriate it notation looks like the following: f(x) B X A where X A means that X is approaching a and the value of the it is B. X 2 X 2 In Example 7, we would express the it as follows: 3 X A X 2
8 B4 Example 8 below: What about more serious holes and gaps in a function? Refer to the graph f(x) This type of function is commonly referred to as a step funcition. What is the it as X approaches 2 of this step function? X f(x) Approaching from the left of X 2 X f(x) Approaching from the right of X 2 As you can see, it depends on whether you approach X 2 from the right or from the left, as to what you would expect the height of the function to be. From the right, it appears that the it would be 2, and from the left, it appears that the it would be. Remember that we said that in order for the it to exist, that it had to be unique? Because of this, the it as X approaches 2, does not exist. Written in appropriate it notation, we get: f(x) Does not exist (DNE) x 2 For completeness, let s write Examples, 2, 4, and 6 in proper it notation. Example : Example 2: Example 4: Example 6: x X 0 x Does not exist x 0 X + X 2 4 x 2 Let s review:. Some its exist and some do not. 2. If the it exists, then it is a real number and it is unique. 3. The proper nototation for a it is: f(x) B x A 4. To evaluate the it, first try substituting A into the function, (i.e., try f(a).) 5. If you cannot evaluate the it by substituting, then try factoring. 6. If neither method works, then graph the function. 7. Finally, the it is simply the anticipated height of f(x) when X A.
9 AA Find the value of y for several values of x and graph each equation. ) y e 2x x y ) y 2 ln x x y Factor. 3) ln 2 x  ln x  2 4) 2 ln 3 x + 3 ln 2 x 5) 3e 2x + 5e x  2 6) e 2x  e x  Solve for x. 7) 2 5x 3 2x 8) 2 x 4 x2 9) e 2x ln 5 Read the information given and answer the questions. The number of bacteria present in a culture is given by N(t) 2000 e.3t. 0) How many bacteria will be in the culture when t 2 hours? ) How many hours would it take for the bacteria count to reach 00,000?
10 AB Find the value of y for several values of x and graph each equation. ) y e 2x x y ) y ln 2x x y Factor. 3) 2ln3 x  8ln x (ln x  2) 4) 2 ln 5 x + 5 ln 4 x 5) 6e 2x + 4e x  2 6) e 2x  ln 2 x Solve for x. 7) e x + 2 x 3 Hint: First multiply both sides 8) 2 5+x 3 2x 9) e 3x ln 4 through by e x, then substitute y for e x. Read the information given and answer the questions. The amount of money which has accumulated after t years at interest rate r under continuous compounding is M Pe rt, where P principle(original money invested), r interest rate, and t time in years. 0) If P $600, r.09, and t 0, compute M. ) If $000 was invested at 5% over 20 years, find M. 2) How long would it take $000 to become $500 if it was invested at 6% interest?
11 AC Find the value of y for several values of x and graph each equation. ) y 2e x x y x y 2) y ln x Factor. 3 3) 3e 2x  7e x + 2 4) 5e 2x  4e x ) ln 2 x  4 ln x +2 6) 2 ln 3 x + 6 ln x Solve for x. 7) 3 2x 4 x2 8) e x  e x 4e x 9) log 4 (ln 5) X Read the information given and answer the questions. The halflife of iodine3 is 8 days. That is the time it takes for one half of the substance to decay or change into another kind of matter. The first question is done as an example. 0) How much of 6 grams of iodine3 will remain after 0 days? The formula needed is the decay formula: Q(t) I e kt where I equals initial amount of the substance, t time, and k decay constant. The decay constant is different for different substances. For Iodine3 it is approximately.087. Q(t) I e kt Q(0) 6(e (.087)(0) ) Q(0) 6(e .87 ) Q(0) 6(.49) 2.5 g remaining after 0 days ) How much iodine would be left after 2 days? 2) How many days would it take for only gram of iodine to remain?
12 AD Find the value of y for several values of x and graph each equation. ) y e x x y 2) y ln x x y Factor. 3) 5e 2x + 3e x  6 e x + 3 4) 6 e 2x  2 e x 5) ln 2 x  ln 2 x + 2 ln x + 6) 3 ln 3 x + 6 ln x Solve for x. 7) e 2x ) ln 2 x  5 ln x ) e 3x ln 2 Read the information given and answer the questions. The halflife of sodium24 is 5 hours. When you are given the halflife af a substance, you can find the decay constant as follows. Start with the decay formula, and plug in the values you know. Q(t) Se kt /2 e k(5) Set the size of the original sample to, and the size of the sample after 5 hours to /2. ln /2 ln e 5k ln /25k ln /25 k.046 0) A 0 gram quantity of sodium24 was studied. How much sodium24 was left after 20 hours? ) How many days would it take for only 2 grams of sodium24 to remain?
13 BA Evaluate the following its from the given graph. ) g(x) 2) r(x) g(x) x 0 r(x) x 2 r(x) x Evaluate the following its by drawing a graph. 3) x r(x) X  2 4) x 2x Evaluate the following its by factoring. 5) X 2 + 3X 0 x 2 X 2 6) x 3 X 2 + X 6 X 2 + 2X 3 Evaluate the following its using any appropriate method. 7) x 2X26 8) x X2 + 2X 9) x π/2 cos X sin 2 X 0) x 0 0 cot 2 θ csc 2 θ s ec 2 θ Hint: Simplify and then evaluate.
14 BB Evaluate the following its from the given graph. ) w(x) 2) q(x) w(x) x 2 q(x) x q(x) x 2 Evaluate the following its by drawing a graph. 3) x x 4) θ π/2 sin θ Evaluate the following its by factoring. 5) 2X 2 6 x 3 X 3 6) X 2 + 9X 20 x 4 X 4 Evaluate the following its using any appropriate method. 7) y 0 log y 8) x 0 2e x (e x ) e 2x e x Hint: Factor, then evaluate. 9) x 2 X 2 0) x 3 X 3 3 X
15 BC Evaluate the following its from the given graph. ) s(x) 2) p(x) s(x) x 2 p(x) x Evaluate the following its by drawing a graph. 3) X 2 X x x 4) θ π/2 cos θ Evaluate the following its by factoring. 5) x X 2 x 6) 3X 7X 2 x 0 X Evaluate the following its using any appropriate method. 7) θ 90 sec 2 θ 3sec 2 θ 8) x π x 9) x 2 X 2 7X + 0 X 2 4 (X + 3) 0) 3 27 x 0 X Hint: Use Pascal s triangle to expand (X+ 3)3.
16 BD Evaluate the following its from the given graph. ) v(x) 2) s(x) 3π/2 π/2 π 2π v(x) x 2 s(x) x π v(x) x 4 Evaluate the following its by drawing a graph. 3) x e x 4) θ 80 2 tan θ Evaluate the following its by factoring. 5) 3X 3 27X x 3 X 3 6) x 4 4 X X 2 6 Evaluate the following its using any appropriate method. 7) x 2 4X 7 + X 8) x π/2 cos x cotx 9) θ 90 sinθ cos θ csc θ + cotθ Hint: Multiply through by the conjugate for  cosθ. 0) x 0 e 2x + e x 2 e x Hint: Factor the numerator.
17 Test A ) The natural logarithm function, ln x, is the inverse of A) sin X B) cos X C) e x D) log x 2) The graph of the e function lies in which quadrants? A) I only B) II only C) I and II D) III only 3) The value of e is approximately A) 27 B) 2.7 C) 270 D).27 9) If the number of bacteria present is given by N (t) 5000 e 3t, how many bacteria will be present when t 5 hours? A) 6,749 B) 22,408 C) 6,600 D).6 x 0 0 0) Use the formula from #9 to find how long it will be until there are 50,000 or more bacteria. A) nearly 0 hours B) nearly 6 hours C) nearly 7 hours D) nearly 8 hours 4) The factorization of ln 2 x  is A) (ln x )(ln x  ) B) (ln x  )(ln x + ) C) 2 ln x (x) D) log x 2A log x B 5) The factorization of 2e 2x + 5e x  2 is A) (2e 2x  3) (e x + 4) B) (2e x + 3) (e x  4) C) (2e x  6) (e x + 2) D) (2e x  3) (e x + 4) ) A triangle has side c 28, m A 34 and m C 72. Use the law of sines to find the length of side a. A) 3.2 B) 6.5 C) 52.8 D) 28 2) What are the reference angle and tan for 405? A) 45, 2 B) 45, C) 60, 3 D) 35, 6) If e 2x 8, then x A).03 B) 2.08 C).69 D).35 3) cos(90  q) csc q A) cos 2 q B) tan q C) sin q cos q D) sin 2 q 7) If e 2+x e ln 3+, then x A). B). C) 4. D) 3. 8) If ln (4e x ) 6, then x A).4 B) 2.4 C) 7.4 D) 4.6 4) A triangle has sides a 8, b 4 and c. Use the law of cosines to solve for angle A. A) 90.7 B) 85.4 C) 33.9 D) ) Change r to rectangular form. 3 cos q  sin q A) 3X  Y 6 B) X  Y 2 C) X  3Y 6 D) 3X 2  Y 2 2
18 Test B ) Which of the following methods can be used to compute a it? I) graphing II) factoring III) the Pythagorean theorem A) III nd IV B) I only 2) Which of the following statements is not true? A) Some its exist. B) A it is a complex number. C) A it is the anticipated height of a function. 3) x 2π cos θ A) 0 B) IV) quadratic formula C) I and II only D) I, II and III D) Limits are unique. C)  D) does not exist 4) X 2 + 3X + 0 x 5 X  5 A) 7 B) does not exist C) 3 D) 7 5) The graph is Y cot θ. 8) 2 ln x + 3 ln x x ln x A) 3 B) 3 C) 0 D) does not exist 9) x 4 X 4  X A) /254 B) 254 C) 0 D) does not exist 0)  x 2X + 7 A) 7 B) 2 C) 0 D) does not exist ) If X + < 2, then the solutions for X are A) X > B) X < 3 C) X > 3 D) no solutions 2) If X + < 7, then the solutions for X are A) X < 48 B) X > 48 C) X < 49 D) X < 6 π 2 cot θ x 2π π 3π 2π 2 A) 00 B) 00 C) 0 D) does not exist 6) e 2x + 2e x x e x A) 0 B) C) 2 D) does not exist cos 7) 2 θ  θ 0 sin θ(sin θ  ) A) 0 B) C)  D) does not exist 3)  B 3 3 A) B2 B) B 3 3 C) B4 D) B8 4) 3(  cos 2 θ) is equal to which of the cos(90  θ) following when θ 90? A) 0 B) 3 C) D) 3 5) B 2 B 3 C 4 C 2 B A) C 6 /B 2 B) B 2 /C 6 C) C 4 B 2 D) C 4 B 2
19 Solutions A  A ) y e2x x y ) y 2 ln x x y ) ln2 x  ln x  2 Y2  Y  2 (Y  2)(Y + ) (ln x  2)(ln x + ) 4) 2 ln3 x + 3 ln2 x 2Y3 + 3Y2 Y2(2Y + 3) ln2 X(2 ln X + 3) 5) 3e2x + 5ex  2 3Y2 + 5Y  2 (3Y  )(Y + 2) (3ex  )(ex + 2) Y 2 6) Y ( Y + ) ( Y ) ( Y ) Y + e x + 0) N t 2000e ) 2 5X 3 2X ln ( 3 2X ) ln 2 5X 5X ln 2 2X ln 3 5X ln 2 2X ln 3 0 X ( 5 ln 2 2 ln 3) 0 X 0 8) 2 X 4 X 2 ln 2 X ln 4 X 2 ( X ) ( ln 2 ) ( X 2 ) ( ln 4 ) X X 2 ln 4 ln 2 X X 2 2 X 2 ( X 2 ) X 2X 4 X 4 X 4 9) e 2X ln 5 ln ln 5 ln e 2X 2X ln ( ln 5 ) ln ( ln 5 ) X.24 2 ) 00,000 2,000e.3t t ln 50.3t ln 50.3 t t 3
20 Solutions A  B ) y e2x x y ) y ln 2x x y 3) 2Y 3 8Y Y 2 2Y ( Y2 4 ) Y 2 4) 2Y 5 + 5Y 4 Y 4 ( 2Y + 5 ) ln 4 X ( 2 lnx + 5 ) Y ( Y + 2 ) ( Y 2 ) 2 lnx ( lnx + 2 ) Y ) 6Y 2 + 4Y 2 2 ( 3Y 2 + 2Y ) 2 ( 3Y ) ( Y + ) 2 ( 3e X ) ( e X + ) 6) e 2X ln 2 X ( e X ln X ) ( e X + ln X ) (difference of 2 squares) 7) e X + 2 X 3 e 2X + 2e 0 3X e 2X 3e X Y 2 3Y ( Y ) ( Y 2 ) 0 ( e X ) e X 2 0 e X ln X ln X 0 e X 2 ln e X ln 2 X ln ) 2 5+X 3 2 X 5 ln 2 + X ln 2 2 ln 3 X ln 3 X ln 2 + X ln 3 2 ln 3 5 ln 2 X ( ln 2 + ln 3) 2 ln 3 5 ln 2 X 2 ln 3 5 ln 2 ln 2 + ln ) e 3X ln 4 ln ln 4 ln e 3X 3X ln ( ln 4 ) ln ( ln 4 ) X. 3 ( 0) M 600e.09 )( 0 ) M 600e.9 M, ( ) M 000e.05 )( 20 ) M 000e M 2, ) M Pe rt e.06t.5 e.06t ln.5 ln e.06t ln.5.06t t ln or about years
21 Solutions A  C ) y 2ex 2) y ln x x y x y ) 3Y 2 7Y + 2 ( 3Y ) ( Y 2 ) ( 3e X ) e X 2 4) 5Y 2 4Y 3 ( 5Y + ) ( Y 3) ( 5e X + ) e X 3 5) Y 2 4 Y + 2 ( Y + 2 ) ( Y 2 ) Y + 2 Y 2 ln X 2 6) 2Y 3 + 6Y 2Y Y ln X ln 2 X + 3 7) 3 2X 4 X 2 ln 3 2X ln 4 X 2 2X ln 3 ( X 2 ) ln 4 2X ln 3 X ln 4 2 ln 4 2X ln 3 X ln 4 2 ln 4 X ( 2 ln 3 ln 4 ) 2 ln 4 X 2 ln 4 2 ln 3 ln ) e X e X 4e X e 2X e 0 4e 0 multiply by e X e 2X 4 e 2X 5 ln e 2X ln 5 2X ln 5 X ( ln 5 / 2 ).8 9) log 4 ( ln 5 ) X 4 X ln 5 ln 4 X ln ( ln 5 ) X ln 4 ln ( ln 5 ) ln ( ln 5 ) X ln ) done ) Q ( t ) Ie kt Q ( 2 ) 2 (.087 ) 6e 6e g 2) Q ( t ) Ie kt 6e.087 ( t ) 6 e.087 ( t ) ln ( / 6 ) ln e.087 ( t ) ln ( / 6 ).087 ( t ) ln ( / 6 ).087 t 20.6 days
22 Solutions A  D ) y ex x y ) y ln x x y ) 5Y 2 + 3Y 6 Y + 3 ( Y + 3) ( 5Y 2 ) Y + 3 5Y 2 5e x 2 4) 6Y 2 2Y 6Y ( Y 2 ) 6e x e x 2 5) Y 2 Y 2 + 2X + ( Y + ) ( Y ) ( Y + ) ( Y + ) Y Y + lnx lnx + 6) 3Y 3 + 6Y 3Y Y ln X ln 2 X + 2 7) Y ( Y + 2 ) ( Y 2 ) 0 Y 2 Y 2 e X 2 e X 2 X ln 2 X ln ( 2 ) X.69 X is undefined 8) Y 2 5Y ( Y 2 ) ( Y 3) 0 Y 2 Y 3 lnx 2 ln X 3 X e 2 X e 3 X 7.39 X ) e 3X ln 2 ln e 3X ln ( ln 2 ) 3X ln ( ln 2 ) ln ( ln 2 ) X.2 3 0) Q ( t ) Se kt.046 ( 20 ) 0e 0e.92 0 (.4 ) 4 g ) Q ( t ) Se kt 2 0e.046 ( t ) 5 e.046t ln ( / 5 ).046t ln ( / 5 ).046 t 35 hours days
23 Sol BA and BB ) g ( X ) 0 X 0 From the left and from the right, the values for g ( X ) get closer to 0 as X approaches 0. 2) r ( X ) DNE (does not exist) X 2 From the left the it 2. From the right the it 0 r ( X ) 0 X 3) ) W ( X ) 0 X 2 2) q ( X ) DNE X 3) q X X 2 3 X X 0 X 2 2 X 4) 4) X 2X 0 X 2 + 3X 0 ( X 2 ) ( X + 5 ) 5) X 2 X 2 X 2 X 2 ( X + 5 ) X 2 6) X 3 X 3 X 2 + X 6 X 2 + 2X 3 ( X 2 ) ( X + 3 ) X 3 ( X + 3) ( X ) ( X 2 ) ( X ) ) X 2X ) X 2 + 2X DNE These numbers will grow X without bound. 9) X π/ 2 cos X sin 2 X cos π / 2 sin 2 π / cos 2 θ cot 2 θ csc 2 θ 0) X 0º sec 2 sin 2 θ sin 2 θ θ X 0º cos 2 θ X 0º cos 2 θ sin 2 θ cos 2 θ X 0º sin 2 θ cos2 θ cos2 θ sin 2 θ X 0º sin 2 θ cos2 θ X 0º cos2 θ cos 2 0º 2X 2 6 5) X 3 X 3 2 ( X 3 ) X 3 X 3 X 2 + 9X 20 6) X 4 X 4 X 4 X 5 X 4 X 4 7) log y log0 Y 0 sin θ sin π / 2 θ π/ 2 2 X 2 9X + 20 X 4 X 4 2e X e X 2e X e X 8) X 0 e 2X e X X 0 e X ( e X ) 9) X 2 X X 3 0) X 3 3 X ( 3 X ) 3 X ( X 5 ) ( 4 5 ) X 4 2
24 Sol BC and B D ) s ( X ) DNE X 2 2) p ( X ) 0 X 3) X 2 X X X ( X ) X X X X X 4) cos θ 0 θ π/ 2 ) V ( X ) DNE X 2 V ( X ) 2 X 4 2) s ( X ) 0 X π 3) 4) DNE X 5) X X 2 X X + 2 X 3X 7X 2 6) X 0 X 3 7 ( 0 ) 3 X ( X ) ( X + ) X X X 0 ( 3 7X ) sec 2 θ 7) θ 90º 3 sec 2 θ tan 2 θ θ 90º 3 sec 2 θ θ 90º sin 2 90º 3 sin 2 θ cos 2 θ 3 cos 2 θ 3 8) X π X DNE X sin 2 θ θ 90º 3 ( X 5 ) X 2 7X + 0 X 2 9) X 2 X 2 4 X 2 ( X + 2 ) X 2 X 5 X 2 X ( X + 3 ) ) X 0 X X 0 ( X 3 + 9X X + 27 ) 27 X X 3 + 9X X X 0 X X 2 + 9X X 0 3X 2 27X 5) X 3 X 3 X 3 3X X + 3 X 3 3 ( 3) ( 3) + 3 6) X 4 X 4 7) X 2 8) X π/ 2 9) θ 90º θ 90º θ 90º θ 90º θ 90º X X X 2 9 X 3 X 3 4 X X 2 6 X 4 ( X + 4 ) X 4 3X ( X + 3) X 3 X X 7 + X 4 ( 2 ) 7 + ( 2 ) 8 3 cos X cot X X π/ 2 cos X cos X sin X X π/2 sin X sin θ cos θ csc θ + cot θ sin θ + cos θ cos θ + cos θ csc θ + cot θ sin θ + sin θ cos θ cos 2 θ csc θ + cot θ sin θ + sin θ cos θ sin 2 θ csc θ + cot θ csc θ + cot θ csc θ + cot θ e X + 2 e 2X + e X 2 e X 0) X 0 e X X 0 e X e X X 0 2 tan θ 2 ( 0 ) 0 θ 80º
25 Test Solutions Test A ) C 2) C 3) B 4) B: y ln x 5) D: y e x 6) A: e 2x 8 y 2  (y + )(y  ) (ln x + )(ln x  ) 2y2 + 5y  2 (2y  3)(y + 4) 2x ln 8 x ln ) A: e 2 +x ln (e ln 3 + ) 2 + x ln 3 + x ln 3  x. 8) C: ln (4ex) ln 4  ln e x ln x ln ) B: N(t) 5000 e 3(5) 5000 e.5 22,408 0) D: 50,000 5,000e.3t 0 e.3 t t ln hours or nearly 8 hours ) where are  5? Test B ) C 2) B 3) B X 2 3X 0 4) D: x 5 X 5 5) D ( X + 2 ) ( ) 7 x 5 ( X + 2 ) ( x 5 ) x 5 ( X 5 ) e 2X + 2e X e X e X + 2 6) C: x e X x e X 2 x e X + 2 sin 2 θ 7) A: θ 0º sin θ ( sin θ ) sin θ θ 0º sin θ 0 θ 0º 0 0 lnx ( 2 lnx + 3 ) 8) A: x lnx x 2 lnx ) A: x 4 X 4 X ) A: (the first term gets closer and x 2X closer to 0) ) D: Absolute value cannot be negative. 2) A: Solving the equality gives us X 48. Substituting 0 for X in the original inequality gives a true result, so the solution is X < ) C: B 3 B 2 B 4 4) B: 3 cos 2 θ cos ( 90º θ) 3 sin2 θ 3 sin θ; 3 sin90º 3 sin θ 5) A B 2 B 3 C 4 C 2 B B C 4 C 2 B C2 4 C B 2 C 6 B 2
ALGEBRA 2/TRIGONOMETRY
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