Section 3.3 Limits Involving Infinity - Asymptotes

Transcription

1 76 Section. Limits Involving Infinity - Asymptotes We begin our discussion with analyzing its as increases or decreases without bound. We will then eplore functions that have its at infinity. Let s consider the following eamples: E. Find a) the and b) the a) In this problem, saying means to let increase without bound. If we make a table of values, we can then see what happens to the function values: The function values are getting closer and closer to zero. Thus,. b) In this problem, saying means to let decrease without bound. If we make a table of values, we can then see what happens to the function values: The function values are getting closer and closer to zero. Thus,. In general, if n is a positive number, then and. n n Let's consider some more eamples to build on what we have learned here:

2 77 E. Find the 6. Here, our function is a constant function so 6 6. E. Find the 6 +. As decreases without bound, so does 6 +. In other words, as gets to be a very large negative number, so does 6 +. In fact, the term will dominate the rest of the epression. So, 6 +. E. 4 Find the As decreases without bound, increases without bound since ( #) 4 + answer. In other words, as gets to be a very large negative number, becomes a very large positive number since the 4 term dominates the epression. So, E. 5 Find the. As increases without bound, decreases without bound since (+ #) answer. So,. E. 6 Find the As increases without bound, both the numerator and the denominator get large. This does not allow us to determine the it. But, since is increasing without

3 78 bound, is staying well away from zero. Thus, we can use the idea that if n is a positive number, then. First, we find the degree of the polynomial n in the denominator. We then multiply the top and bottom of the rational epression by / raised to that power. In this eample, we will multiply the top and bottom by. After that, we can take the it: Thus, E. 7 Find 4 Since the degree of the polynomial in the denominator is 4, will we multiply top and bottom by and take the it: 4 + E. 8 Find the the 5 term in the numerator decreases without bound as. Thus, since

4 79 We can also have its at infinity as approaches a number. Consider the following eample: E. 9 Find the (+ ). First consider the left-hand it. As, ( + ) becomes a smaller and smaller positive number. Thus, (+ ) becomes a larger and larger positive number or increases without bound. Hence, (+ ). Now consider the right-hand it. As +, ( + ) becomes a smaller and smaller positive number. Thus, (+ ) becomes a larger and larger positive number or increases without bound. Hence, Since, (+ ). E. Find the (+ ) (+ ), then (+ ). First consider the left-hand it. As, + becomes a smaller and smaller negative number. Thus, + becomes a larger and larger negative number or decreases without bound. Hence, +. Now consider the right-hand it. As +, + becomes a smaller and smaller positive number. Thus, + becomes a larger and larger positive number or increases without bound. Hence, + +. Since,

5 , then + does not eist. We now can use the ideas of its involving infinity to discuss the asymptotes of graphs. The asymptotes of a graph are the straight lines that a function approimates as the values of increase without bound, decrease without bound, or approach a number. We will compare our definition of asymptotes to one that you may have seen in College Algebra. The definitions in Calculus are much broader and work with a wider range of functions. It is very important to be sure that f() is in lowest terms. Vertical Asymptotes: College Algebra: a is a vertical asymptote if f(a) Non Zero # Calculus: If a. f() ± and/or a is a vertical asymptote. f() ±, then + a Find the vertical asymptotes of the following: E. f() f() E. g(). ( has a vertical asymptote of since (also 6+ 5)( ) +. + ). We first have to check to see if it is reduced to lowest terms: ( Since 6+ 5)( ) 5 ± ( 5)( ) ( ) ( 5)( )( ) ± and ( 5)( ) ± ( 5)( ). ±, g has vertical asymptotes of 5 and. Since divided out, the function has a hole at.

6 8 In the last eample, the function had vertical asymptotes of 5 and. This means that as get close to either one or five, the function approimates the vertical line. To see how this works, let s look at the graph of g: The graph approimates the line. The graph approimates the line 5. The graph has a hole at. Horizontal Asymptotes: College Algebra: ) If the degree of the polynomial in the denominator is equal to the degree of the polynomial in the numerator, then y a is a horizontal asymptote where a and b are b the leading coefficients of the polynomials in the numerator and denominator respectively. ) If the degree of the polynomial in the denominator is greater than the degree of the polynomial in the numerator, then y is a horizontal asymptote. ) If the degree in the polynomial in the numerator is larger than the degree of the polynomial in the denominator, then there is no horizontal asymptote.

7 8 Calculus: If f() R, and/or f() R, where R ±, then y R is a horizontal asymptote. Find the Horizontal Asymptotes of the following: 6 E. h(). Since h has a horizontal asymptote of y 6 7. Also, E. 4 f() Since (and 4 + has a horizontal asymptote of y. 6 7, then ), then f In the last eample, the function had a horizontal asymptote of y. This means that as get increases or decreases without bound, the function approimates the horizontal line y. To see how this works, let s look at the graph of f:

8 8 The graph approimates the line y The graph approimates the line y Slant (Oblique) asymptotes: College Algebra: If the degree of the polynomial in the numerator is one more than the degree of the polynomial in the denominator, then divide. The result is f() m + b + h(). Thus, y m + b is a slant asymptote. Calculus: If f() m + b + h() where then y m + b is the slant asymptote. h(), and/or h(), Note: A function cannot have both a horizontal and a slant asymptote. Find the slant asymptote of the following: E. 5 y 5+ We will use synthetic division to divide: Thus, y Since 4

9 84 ( ) then the function has a slant asymptote of y., In the last eample, the function had a slant asymptote of y. This means that as get increases or decreases without bound, the function approimates the line y. To see how this works, let s look at the graph of function with the line y drawn as a dashed line: The graph approimates the line y The graph approimates the line y Now, we will put all the techniques that we have learned thus far in this chapter together and outline a general procedure that we will follow when graphing functions. The procedure is outlined on the net page and afterwards, we will use it to graph some eamples.

10 Steps for Graphing f() ) Find the domain of f. Reduce f() to lowest terms and identify any "holes" in the function. ) Find the y-intercept and, if they are relatively easy to find, the -intercept(s). ) Find f '() and f "(). 4) Find all asymptotes of f(): I) Vertical Asymptotes: a is a vertical asymptote if f() ± and/or f() ±. a + a II) III) Horizontal Asymptotes: y b is a horizontal asymptote if f() b and/or f() b. Slant Asymptotes: y m + b is a slant asymptote if f() m + b + g() where g() and/or g(). 5) Find all the critical points, i.e., all the values c of the domain of f where f '(c) and undefined. Determine where the function is increasing (f '() > ) or decreasing (f '() < ). 6) Use the second the derivative test to verify which of the critical points are relative maimums and minimums (If f "(c) >, minimum. If f "(c) <, maimum.) If the test fails (f "(c) or is undefined) use the first derivative test: I) If f '() < to the left of c and f '() > to the right of c, then f(c) is a relative minimum. II) If f '() > to the left of c and f '() < to the right of c, then f(c) is a relative maimum. 7) Find all the inflection points, i.e., all the values d of the domain of f where f "(d) and undefined. Determine the concavity (If f "() <, down. If f "() >, up). 8) Sketch the graph. Feel free to plot some additional point 85

11 86 Graph the following: E. 6 f() I) Since f is a polynomial, the domain is (, ). II) Since f() () 4 6() + 8(), the y-intercept is (, ). Setting f() and solving yields: ( 6 + 8) or or 6± 8± ( 6) 4()(8) ().6 or.7. Thus, the -intercepts are (, ), (.6, ), and (.7, ). III) f '() d d [ ] f "() d d [ ] IV) f() is a polynomial so it has no asymptotes. V) f '() is a polynomial so it is defined for all real numbers. Setting f '() and solving yields: ( 4 + ) ( )( ),, and Thus,,, and are the critical values. Evaluating f at these values yields: f() () 4 6() + 8() f() () 4 6() + 8() 5 f() () 4 6() + 8() 7 So, (, ), (, 5), and (, 7) are the critical points. Marking,, and on the number line, we can find where f is increasing and decreasing:

12 87 Pick Pick.5 Pick Pick 4 f '( ) f '(.5) f '() f '(4) Thus, f is increasing on (, ) U (, ) and decreasing on (, ) U (, ). VI) f "() 6() 96() so (, ) is a relative minimum (verifies the results in part V). f "() 6() 96() so (, 5) is a relative maimum (verifies the results in part V). f "() 6() 96() so (, 7) is a relative minimum (verifies the results in part V). VII) f "() is a polynomial so it is defined for all real numbers. Setting f "() and solving yields: ( 8 + ) 8± ( 8) 4()() () 4± 7.45 or.. Evaluating f() at these values yields: f(.45). and f(.).6. Thus, (.45,.) and (.,.6) are the possible inflection points. Marking.45 and. on the number line, we can find where f is concave up or down:.45. Pick Pick Pick f "() 6 f "() 4 f "() 7 Thus, f is concave up on (,.45) U (., ) and concave down on (.45,.).

13 88 VIII) Now use the information from parts I - VII to sketch the graph of f: E. 7 f() + I) Since +, then the domain of f is (, ). II) Since f() () () + Setting f() and solving yields: (multiply by + + ) ( + ) + ( + ). So, the only -intercept is (, )., the y-intercept is (, ).

14 89 III) f '() d d [ ] + ( + ) ( + )[] ( [] + ) f "() d d [ + ] ( + ) ( + ) d d + ( ( + ) d [ + ] ( d 4 ( + ) ( + ) [] ( + 4 ( ( + ) [] + ( 4 ( + ) ( + )[( + ) [] 4 ( + ) ( + )[ 4 [ ( 6] + ) ( + ) ( ) + ) + ) ) [( ) ( + ( ( + ) ) 4]. d d [] [( d d [ + ] +. ( + ) + ) d d + )] ( + ) () )()] ] + ) IV) The function has no vertical asymptotes. But, ± + ± + ± + Thus, f has a horizontal asymptote of y. +. V) Since +, f '() is defined for all real numbers. Setting f '() and solving yields: + ( + ) (multiply by ( + ) ) + ( + ) ( + ) + ( ) ( )( + ) and. ( + )

15 9 Evaluating f at these values yields: ( ) f( ) ( ) + and f() () () +. So, (, ) and (, ) are the critical points. Marking and on the number line, we can find where f is increasing and decreasing: Pick Pick Pick f '( ). f '() f '(). Thus, f is increasing (, ) and decreasing on (, ) U (, ). ( )(( ) ) VI) f "( ) (( ) + ) 4 8 so (, ) is a relative minimum (verifies the results in part V). ()(() ) f "() (() + ) 4 8 so (, ) is a relative maimum (verifies the results in part V). VII) Since +, then f "() is defined for all real numbers. Setting f "() and solving yields: ( ) ( + ) ( + ) (multiply by ( + ) ) ( ) ( + ) ( ) or ( + ) or ± ±.7. Evaluating f at these values yields: f(.7).4, f(), and f(.7).4. Thus, (.7,.4), (, ), (.7,.4) are the possible inflection points. Marking these values on the number line, we can find the concavity of f:

16 9.7.7 Pick Pick Pick Pick f "( ) f "( ) f "() f "() Hence, f is concave up on (.7, ) U (.7, ) and concave down on (,.7) U (,.7). VIII) Now, we can sketch the graph:.5.5 E. 8 f() + I) Since f is a rational function, it is undefined when the denominator is or when. Thus, the domain is (, ) U (, ). II) Since f() is undefined, there is no y-intercept. Setting f() and solving yields: (multiply by ) Thus, (, ) is the - intercept.

17 9 III) First rewrite f() as Thus, f '() d d [ + ] and f "() d d [ ]. IV) Since + ± of. Also, since + ± horizontal asymptote of y. ±, f has a vertical asymptote +, f has a V) f '() is undefined at, but is not in the domain of f. Thus, is not a critical value. Setting f '() and solving yields: (multiply by ) ( ), no solution. Thus, f has no critical points. Hence, we mark only the value where f is undefined on the number line and find where f is increasing and decreasing: Pick Pick f '( ) f '() Thus, f is decreasing on (, ) U (, ) and increasing nowhere. VI) Since there are no critical points, this part is not applicable. VII) f "() is undefined at, but is not in the domain of f. Thus, is not to be considered. Setting f "() and solving yields:

18 9 (multiply by ) ( ), no solution. Thus, f has no inflection points. Hence, we mark only the value where f is undefined on the number line and determine the concavity of f: Pick Pick f "( ) f "() Thus, f is concave up on (, ) and concave down on (, ). VIII) Now, we will sketch the graph:

19 94 E. 9 f() + 9 I) Since f is a rational function, it is undefined when the denominator is or when. Thus, the domain is (, ) U (, ). II) III) Since f() is undefined, there is no y-intercept. Setting f() and solving yields: + 9 (multiply by ) ± 9, no real solution. Thus, there are no - intercepts. First rewrite f() as Thus, f '() d d [ + 9 ] 9 9 and f "() d d [ 9 ] 8 8. IV) Since ± + 9 of. Also, since 9 ± asymptote of y. ±, f has a vertical asymptote, f has a slant V) f '() is undefined at, but is not in the domain of f. Thus, is not a critical value. Setting f '() and solving yields: 9 (multiply by ) 9 9 ( )( + ) or. Thus, and are the critical values.

20 95 Evaluating f at these values yields: f() () + 9 () 6 and f( ) ( ) ( ) Thus, (, 6) and (, 6) are the critical points. Marking these values and the value where f is undefined on the number line, we can find where f is increasing and decreasing: Pick 4 Pick Pick Pick 4 f '( 4) f '( ) f '() f '(4) Thus, f is increasing on (, ) U (, ) and decreasing on (, ) U (, ). 8 VI) f "( ) so (, 6) is a relative ( ) maimum (verifies the results in part V). f "() 8 so (, 6) is a relative minimum () (verifies the results in part V). VII) f "() is undefined at, but is not in the domain of f. Thus, is not to be considered. Setting f "() and solving yields: 8 (multiply by ) ( 8 ) 8, no solution. Thus, f has no inflection points. Hence, we mark only the value where f is undefined on the number line and determine the concavity of f:

21 96 Pick Pick f "( ) 8 f "() 8 Thus, f is concave up on (, ) and concave down on (, ). VIII) Now, we will sketch the graph: E. f() +. I) Since +, then the domain of f is (, ). II) Since f() () + Setting f() and solving yields:, the y-intercept is (, ).

22 97 (multiply by + + ) ( + ) + ( + ), no solution. So, there are no -intercepts. III) f '() d d [ + ] d d [( + ) ] ( + ) d d [ + ] ( + ) [] ( ( + ). ( f "() d d [ ] + ) + ) ( + ) d [] () d 4 ( + ) d d [( + ) ] ( + ) [] () [( + )] d ( + ) d 4 ( + ) ( + ) [] + 4 ( + ) () 4 ( + ) ( + )[( + ) [] + 4()] 4 ( + ) ( + )[ 6+ 8 ] 4 ( + ) [6 6] ( + ) 6( ) ( + ). IV) The function has no vertical asymptotes. But, ± + ± + ± + Thus, f has a horizontal asymptote of y. +. V) Since +, f '() is defined for all real numbers. Setting f '() and solving yields:

23 98 ( ) + ( + ) (multiply by ( + ) ) ( + ) ( + ) Evaluating f at this value yields: f() () +. So, (, ) is the critical point. Marking on the number line, we can find where f is increasing and decreasing: Pick Pick f '( ).5 f '().5 Thus, f is increasing (, ) and decreasing on (, ). 6(() ) VI) f "() (() + ) so (, ) is a relative maimum (verifies the results in part V). VII) Since +, then f "() is defined for all real numbers. Setting f "() and solving yields: 6( ) ( + ) ( + ) (multiply by ( + ) ) 6( ) ( + ) ( + ) 6( ) 6( )( + ) or. Evaluating f at these values yields: f( ).5, and f().5. Thus, (,.5) and (,.5) are the possible

24 99 inflection points. Marking these values on the number line, we can find the concavity of f: Pick Pick Pick f "( ) f "() f "() Hence, f is concave up on (, ) U (, ) and concave down on (, ). VIII) Now, we can sketch the graph: E. f() 6. I) Since 6 ( 4)( + 4), then f is undefined at 4 and 4. Thus, the domain of f is (, 4) U ( 4, 4) U (4, ). II) Since f() () () 6 6 (, ). Setting f() and solving yields: 6 ( 6) (multiply by 6) 6 ( 6), the y-intercept is

25 So, the -intercept is (, ). III) f '() d d [ ] 6 ( 6) ( 6) [4] ( 6) ( 64 6) [] f "() d d [ ] ( 6) d [ 64] d 4 ( ( 6) ( 64) 6) 6) [ 64] 4 ( d d [ ) d d ( [( 6) ] 4 6) ( ( 64) [( 6)] 6) [ 64] + 4 ( 8 ( 6) () ( 6) 6)[( 4 ( 6) [ 64] + 8()] ( 6) 6)[ ( [9 ( ( 4] 6) 6) ( 6) 64 ( + 6) ]. d d ] ( d d [ 6] ( 6) 64 6) IV) Since ± and ±, ± 4 6 ± 4 6 f has two vertical asymptotes at 4 and 4. Also, ± 6 y. ± 6 ± 6. Thus, f has a horizontal asymptote of V) f '() is undefined at ± 4, but ± 4 are not in the domain of f. Thus, these are not critical values.

26 Setting f '() and solving yields: ( 64 6) ( 6) (multiply by ( 6) ) ( 64 6) ( 6) 64 Evaluating f at this value yields: f() () () 6 6, So, (, ) is the critical point. Marking and the values where f is undefined on the number line, we can find where f is increasing and decreasing: 4 4 Pick 5 Pick Pick Pick 5 f '( 5) f '( ) f '() f '(5) Thus, f is increasing (, 4) U ( 4, ) and decreasing on (, 4) U (4, ). VI) f "() (() 64 (() + 6) 6) so (, ) is a relative maimum (verifies the results in part V). VII) f '() is undefined at ± 4, but ± 4 are not in the domain of f. Thus, these values are not to be considered. Setting f '() and solving yields: ( 64 ( + 6) 6) ( 6) (multiply by ( 6) ) ( 64 ( + 6) 64( + 6) + 6 6) ( 6)

27 6 ±, no solution. Thus, there are no inflection points. Marking only values that make f undefined on the number line, we can find the concavity of f: 4 4 Pick 5 Pick Pick 5 f "( 5) f "() f "(5) Hence, f is concave up on (, 4) U (4, ) and concave down on ( 4, 4). VIII) Now, we can sketch the graph: