Section 3.2 Drawing graphs using first-derivative tests

Transcription

1 Section Drawing graphs using first-derivative tests Overview: In this section we sketch graphs of functions constructed from powers, linear combinations, products, and quotients b studing formulas for the functions and then appling procedures from Section 1 to analze their first derivatives Topics: Sketching graphs of polnomials Graphs of polnomials near = 0 Sketching graphs of rational functions Functions with restricted domains Sketching graphs of polnomials As we saw in Section 1, the its of a polnomial as ± are determined b its highest degree term This is because, when is large, the lower degree terms are relativel small compared to the highest-degree term The polnomial = 1 +0, for eample, has the same its as ± as its term of highest degree 1 : (1 + 0) = (1 ) = (1 + 0) = (1 ) = (1) This is because, for 0, ( = ) and the epression in parentheses on the right is approimatel equal to 1 negative To determine the overall shape of the graph, we stud its first derivative Eample 1 for large positive or Sketch the graph of = Show where the function is increasing and decreasing and plot an local or global maima and minima The polnomial = is continuous on (, ) Its derivative () = d d (1 + 0) = 4 = ( 4) () is zero at = 0 and = 4 and has constant sign on each of the open intervals (,0),(0,4), and (4, ) set off b the zeros We can determine the signs b calculating the value of the derivative at one point in each interval, and use this information with Theorem of Section 1 to see where the function is increasing and decreasing : ( 1) = ( 1)( 1 4) = 5 > 0 = () > 0 and on (,0) (1) = (1)(1 4) = < 0 = () < 0 and on (0,4) (5) = (5)(5 4) = 5 > 0 = () > 0 and on (4, ) If we allow nonopen intervals in appling Theorem of Section 1, we would conclude that the function = of Eample 1 is, in fact, increasing on the closed intervals (, 0] and [4, ) and is decreasing on the closed interval [0, 4] We onl consider open intervals in the solutions of eamples, tune-up problems, and basic eercises in this section, but consider nonopen intervals in the solutions of some eplorator eercises 59

2 p 60 Drawing graphs using first-derivative tests Section We displa this information with an -ais in Figure 1 We use it with the facts, established above, that the function tends to as and tends to as, and plot the values ( ) = 1 ( ) ( ) + 0 = 7,(0) = 0, (4) = 1 (4 ) (4 ) + 0 = 9 1, and (6) = 1 (6 ) (6 ) + 0 = 0 to obtain the graph in Figure The function has a local maimum of 0 at = 0, and a local minimum of 9 1 at = 4 It has no global maima or minima 0 > 0 < 0 > FIGURE 1 FIGURE Question 1 Use the factorization () to eplain wh the derivative changes sign (a) at = 0 and (b) at = 4 Graphs of polnomials near = 0 Just as we can determine the its of a polnomial as ± b looking at its term of highest degree, we can anticipate that its graph near = 0 will look like the graph of its lowest degree term or terms We use the following result Theorem 1 The graph of a polnomial = a + b n 0 + [terms of degree > n 0 ] with n 0 a positive integer and b 0 is closel approimated b the graph = a + b n 0 of its lowest-order term or terms for sufficientl close to 0 The approimating polnomial = a+b n 0 in Theorem 1 has two terms if a 0 and one term if a = 0 This result is a consequence of the fact that for ver small, is relativel small compared to, is relativel small compared to, 4 is relativel small compared to, and so forth (see Eercises 1 and ) The theorem is illustrated in Figure, which shows the graph of the polnomial = of Eample 1 with the parabola = + 0 that is the graph of its lowest order terms The curve = is closel approimated b the parabola = + 0 near = 0 because = ( 1 ) + 0 and for small, 1 is close to 0 = and = + 0 FIGURE 4 6

3 Section Drawing graphs using first-derivative tests, p 61 Drawing graphs of rational functions Recall that the rational functions are constructed from ponomials b taking linear combinations, products, and quotients To sketch the graph of a rational function, we first learn what we can b studing its formula This often involves finding its its as ±, studing its behavior near an vertical asmptotes, and noting whether the function is even or odd In some cases, we can also learn something about the graph for small b appling Theorem 1 to polnomials that are parts of the formula for the rational function Then we stud the function s derivative Eample Use the formulas for the function and its first derivative to sketch the graph of = /( ) Studing the function: Recall that the its as ± of a quotient of two polnomials is the it of the quotient of their terms of highest degree Therefore, = = = = = = This can be shown b dividing the numerator or denominator b n where n is the lower of the degrees of the numerator and denominator In this case, we divide the numerator and denominator b to obtain for 0, = = 1 / () This shows that as and as since 1 / is close to 1 for large positive or negative Equation () also shows that the graph looks like the line = for large positive and negative since 1 / 1 for large The function = /( ) can change sign either at = 0, where its numerator is zero and the function is zero, or at =, where the denominator is zero and the graph has a vertical asmptote These two points set off three open intervals (,0),(0,), and (, ) on which the function is continuous and nonzero has therefore has constant sign We could find the signs b calculating sample values Instead, we determine them in this case b finding the signs of the numerator and denominator, as in the following table = < 0 > 0 < 0 < 0 0 < < > 0 < 0 < 0 > > 0 > 0 > 0 The information we have obtained about the function is displaed in Figure 4, where we have written above to indicate that the graph has a vertical asmptote at = Then we can see from Figure 4 that tends to as 0 + and tends to as 0 because > 0 just to the right of 0 and < 0 just to the left of 0

4 p 6 Drawing graphs using first-derivative tests Section = 0 FIGURE 4 < 0 0 < 0 > 0 Studing the derivative: The Quotient Rule gives the formula = d ( ) ( ) d d ( ) d ( ) d = d ( ) = ( )() (1) ( ) = 6 ( ) = 6 ( 6) = ( ) ( ) (4) The derivative (4) can change sign at = 0 and = 6, where the numerator is zero and the function has critical points, or at =, where the denominator is zero and the graph has a vertical asmptote We could find the signs of the derivative b studing the factors and of its numerator and its denominator This time, we calculate sample values instead: ( 1) = () = (4) = (7) = (1)(1 6) (1 ) = 7 16 > 0 = > 0 and in (,0) ()( 6) ( ) = 8 < 0 = < 0 and in (0,) (4)(4 6) (4 ) = 8 < 0 = < 0 and in (,6) (7)(7 6) (7 ) = 7 16 > 0 = > 0 and in (6, ) This information is displaed in Figure 5 > 0 < 0 < 0 > 0 FIGURE Drawing the graph: To obtain the graph of the function in Figure 6, we draw the asmptote = and have the curve go up on its right side and down on its left side We plot the values ( 6) = ( 6) /( 6 ) = 6/( 9) = 4,(0) = 0, (6) = 6 /(6 ) = 6/ = 1, and (9) = 9 /(9 ) = 81/6 = 1 1, and use the information in Figures 4 and 5 The function has a local maimum of 0 at = 0, and a local minimum of 1 at = 6 Notice that the graph looks like = 1 near = 0 This is because the term in the denominator of = /( ) is small relative to for small

5 Section Drawing graphs using first-derivative tests, p 6 0 = FIGURE Question Use formula (4) to eplain wh the derivative in Eample changes sign at = 0 and at = 6, but not at = Eample Sketch the curve = /( + 1) b analzing the formulas for the function and its first derivative Studing the function: The function = /( + 1) is even and is positive for all Its its as ± are the same as the its of the highest power terms in its numerator and denominator: ± + 1 = ± = 1 = 1 ± We could also find these its b dividing the numerator and denominator b : For 0, = + 1 = / This formula shows that /( + 1) 1 as ±, as we concluded above, so that the line = 1 is a horizontal asmptote of the graph Studing the derivative: B the Quotient Rule, () = d ( ) ( + 1) d d ( ) d d ( + 1) d = + 1 ( + 1) = ( + 1)() () ( + 1) = ( + 1) Since the denominator is positive for all, the derivative is positive and the function is increasing on (0, ) and the derivative is negative and the function is decreasing on (,0) Drawing the graph: The graph is smmetric about the -ais because the function is even To draw the graph in Figure 7, we use the information obtained above and plot the values (0) = 0 /(1 + 0 ) = 1 and (±) = /(1 + ) = 4 Notice that 5 the graph looks like = near = 0 This because the in the denominator of = /( + 1) is small relative to the 1 for small

6 p 64 Drawing graphs using first-derivative tests Section = FIGURE 7 Functions with restricted domains If a function s formula involves even roots or irrational powers, then it is not defined where the epressions whose roots or powers are taken are negative Eample 4 Sketch the graph of h() = b analzing the formulas for the function and its first derivative Studing the function: h() = is even and is defined, nonnegative, and continuous where, which is on the intervals (, ] and [, ) Also ± h() = because as ± Studing the derivative: B the rule for finding derivatives of powers of functions, h () = d d [( ) 1/ ] = 1 ( ) 1/ d d ( ) = 1 (5) ( ) = 1/ The derivative is defined for < and for > It is negative for < and positive for >, so the function is decreasing on (, ) and increasing on (, ) We plot the values h(± ) = 0 and h(±4) = 4 = 1 = 6 to draw its graph in Figure 8 The graph is smmetric about the -ais because the function is even = h() 4 h() = FIGURE Question Show that the tangent line at to the graph of h() = approaches the vertical line = as approaches from the right and approaches the vertical line = as approaches from the left Question 4 The slope (5) of the tangent line at to the graph of h() = in Figure 8 tends to 1 as and tends to 1 as Illustrate this b calculating

7 Section Drawing graphs using first-derivative tests, p 65 Responses Response 1 the approimate decimal values of the slope (a) at = 5,10, and 50 and (b) at = 5, 10, and 50 (a) The derivative = 1 (4 ) changes sign at = 0 because the factor changes sign and the factor 4 does not (b) The derivative = 1 (4 ) changes sign at = 4 because the factor 4 changes sign and the factor does not Response (a) The derivative ( 6) = changes sign at = 0 because changes sign ( ) but 6 and ( ) do not The derivative changes sign at = 6 because 6 changes sign but and ( ) do not The derivative does not change at = because, 6 and ( ) do not change sign there Response The tangent line at to the graph = h() in Figure 8 approaches the vertical line = as approaches from the right because the derivative h () = / tends to The tangent line at approaches the vertical line = as approaches from the left because the derivative h () = / tends to Response 4 (a) h (5) = 5 5 h (50) = (b) h ( 5) = 5 5 h (50) = = h (10) = = = h ( 10) = = = 1015 = 1015