# Curve Sketching. 96 Chapter 5 Curve Sketching

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1 96 Chpter 5 Curve Sketching 5 Curve Sketching A B A B A Figure 51 Some locl mximum points (A) nd minimum points (B) If (x, f(x)) is point where f(x) reches locl mximum or minimum, nd if the derivtive of f exists t x, then the grph hs tngent line nd the tngent line must be horizontl This is importnt enough to stte s theorem: Whether we re interested in function s purely mthemticl object or in connection with some ppliction to the rel world, it is often useful to know wht the grph of the function looks like We cn obtin good picture of the grph using certin crucil informtion provided by derivtives of the function nd certin limits A locl mximum point on function is point (x, y) on the grph of the function whose y coordinte is lrger thn ll other y coordintes on the grph t points close to (x, y) More precisely, (x, f(x)) is locl mximum if there is n intervl (, b) with < x < b nd f(x) f(z) for every z in (, b) Similrly, (x, y) is locl minimum point if it hs loclly the smllest y coordinte Agin being more precise: (x, f(x)) is locl minimum if there is n intervl (, b) with < x < b nd f(x) f(z) for every z in (, b) A locl extremum is either locl minimum or locl mximum Locl mximum nd minimum points re quite distinctive on the grph of function, º½ Å Ü Ñ Ò Å Ò Ñ nd re therefore useful in understnding the shpe of the grph In mny pplied problems we wnt to find the lrgest or smllest vlue tht function chieves (for exmple, we might wnt to find the minimum cost t which some tsk cn be performed) nd so identifying mximum nd minimum points will be useful for pplied problems s well Some exmples of locl mximum nd minimum points re shown in figure 51 THEOREM 51 Fermt s Theorem If f(x) hs locl extremum t x = nd f is differentible t, then f () = 0 Thus, the only points t which function cn hve locl mximum or minimum re points t which the derivtive is zero, s in the left hnd grph in figure 51, or the derivtive is undefined, s in the right hnd grph Any vlue of x for which f (x) is zero or undefined is clled criticl vlue for f When looking for locl mximum nd minimum points, you re likely to mke two sorts of mistkes: You my forget tht mximum or minimum cn occur where the derivtive does not exist, nd so forget to check whether the derivtive exists everywhere You might lso ssume tht ny plce tht the derivtive is zero is locl mximum or minimum point, but this is not true A portion of the grph of f(x) = x 3 is shown in figure 52 The derivtive of f is f (x) = 3x 2, nd f (0) = 0, but there is neither mximum nor minimum t (0, 0) Since the derivtive is zero or undefined t both locl mximum nd locl minimum points, we need wy to determine which, if either, ctully occurs The most elementry pproch, but one tht is often tedious or difficult, is to test directly whether the y coordintes ner the potentil mximum or minimum re bove or below the y coordinte t the point of interest Of course, there re too mny points ner the point to test, but little thought shows we need only test two provided we know tht f is continuous (recll tht this mens tht the grph of f hs no jumps or gps) Suppose, for exmple, tht we hve identified three points t which f is zero or nonexistent: (x 1, y 1 ), (x 2, y 2 ), (x 3, y 3 ), nd x 1 < x 2 < x 3 (see figure 53) Suppose tht 95

2 51 Mxim nd Minim Chpter 5 Curve Sketching close together Nevertheless, becuse this method is conceptully simple nd sometimes esy to perform, you should lwys consider it Figure 52 No mximum or minimum even though the derivtive is zero we compute the vlue of f() for x 1 < < x 2, nd tht f() < f(x 2 ) Wht cn we sy bout the grph between nd x 2? Could there be point (b, f(b)), < b < x 2 with f(b) > f(x 2 )? No: if there were, the grph would go up from (, f()) to (b, f(b)) then down to (x 2, f(x 2 )) nd somewhere in between would hve locl mximum point But t tht locl mximum point the derivtive of f would be zero or nonexistent, yet we lredy know tht the derivtive is zero or nonexistent only t x 1, x 2, nd x 3 The upshot is tht one computtion tells us tht (x 2, f(x 2 )) hs the lrgest y coordinte of ny point on the grph ner x 2 nd to the left of x 2 We cn perform the sme test on the right If we find tht on both sides of x 2 the vlues re smller, then there must be locl mximum t (x 2, f(x 2 )); if we find tht on both sides of x 2 the vlues re lrger, then there must be locl minimum t (x 2, f(x 2 )); if we find one of ech, then there is neither locl mximum or minimum t x 2 EXAMPLE 52 Find ll locl mximum nd minimum points for the function f(x) = x 3 x The derivtive is f (x) = 3x 2 1 This is defined everywhere nd is zero t x = ± 3/3 Looking first t x = 3/3, we see tht f( 3/3) = 2 3/9 Now we test two points on either side of x = 3/3, mking sure tht neither is frther wy thn the nerest criticl vlue; since 3 < 3, 3/3 < 1 nd we cn use x = 0 nd x = 1 Since f(0) = 0 > 2 3/9 nd f(1) = 0 > 2 3/9, there must be locl minimum t x = 3/3 For x = 3/3, we see tht f( 3/3) = 2 3/9 This time we cn use x = 0 nd x = 1, nd we find tht f( 1) = f(0) = 0 < 2 3/9, so there must be locl mximum t x = 3/3 Of course this exmple is mde very simple by our choice of points to test, nmely x = 1, 0, 1 We could hve used other vlues, sy 5/4, 1/3, nd 3/4, but this would hve mde the clcultions considerbly more tedious EXAMPLE 53 Find ll locl mximum nd minimum points for f(x) = sin x + cos x The derivtive is f (x) = cos x sin x This is lwys defined nd is zero whenever cos x = sin x Reclling tht the cos x nd sinx re the x nd y coordintes of points on unit circle, we see tht cos x = sin x when x is π/4, π/4 ±π, π/4 ±2π, π/4 ±3π, etc Since both sine nd cosine hve period of 2π, we need only determine the sttus of x = π/4 nd x = 5π/4 We cn use 0 nd π/2 to test the criticl vlue x = π/4 We find tht f(π/4) = 2, f(0) = 1 < 2 nd f(π/2) = 1, so there is locl mximum when x = π/4 nd lso when x = π/4 ± 2π, π/4 ± 4π, etc We use π nd 2π to test the criticl vlue x = 5π/4 The relevnt vlues re f(5π/4) = 2, f(π) = 1 > 2, f(2π) = 1 > 2, so there is locl minimum t x = 5π/4, 5π/4 ± 2π, 5π/4 ± 4π, etc x 1 b x 2 x 3 Figure 53 Testing for mximum or minimum It is not lwys esy to compute the vlue of function t prticulr point The tsk is mde esier by the vilbility of clcultors nd computers, but they hve their own drwbcks they do not lwys llow us to distinguish between vlues tht re very

3 52 The first derivtive test Chpter 5 Curve Sketching Exercises 51 In problems 1 12, find ll locl mximum nd minimum points (x, y) by the method of this section 1 y = x 2 x 2 y = 2 + 3x x 3 3 y = x 3 9x x 4 y = x 4 2x y = 3x 4 4x 3 6 y = (x 2 1)/x 7 y = 3x 2 (1/x 2 ) 8 y = cos(2x) x j x 1 x < 2 9 f(x) = x 2 x 2 8 < x 3 x < 3 10 f(x) = x 3 3 x 5 : 1/x x > 5 11 f(x) = x 2 98x + 4 j 2 x = 0 12 f(x) = 1/x 2 x 0 13 Recll tht for ny rel number x there is unique integer n such tht n x < n + 1, nd the gretest integer function is given by x = n, s shown in figure 31 Where re the criticl vlues of the gretest integer function? Which re locl mxim nd which re locl minim? 14 Explin why the function f(x) = 1/x hs no locl mxim or minim 15 How mny criticl points cn qudrtic polynomil function hve? 16 Show tht cubic polynomil cn hve t most two criticl points Give exmples to show tht cubic polynomil cn hve zero, one, or two criticl points 17 Explore the fmily of functions f(x) = x 3 + cx + 1 where c is constnt How mny nd wht types of locl extremes re there? Your nswer should depend on the vlue of c, tht is, different vlues of c will give different nswers 18 We generlize the preceding two questions Let n be positive integer nd let f be polynomil of degree n How mny criticl points cn f hve? (Hint: Recll the Fundmentl Theorem of Algebr, which sys tht polynomil of degree n hs t most n roots) The method of the previous section for deciding whether there is locl mximum or minimum t criticl vlue is not lwys convenient We cn insted use informtion bout the derivtive f (x) to decide; since we hve lredy hd to compute the derivtive to find the criticl vlues, there is often reltively little extr work involved in this method How cn the derivtive tell us whether there is mximum, minimum, or neither t point? Suppose tht f () = 0 If there is locl mximum when x =, the function must be lower ner x = thn it is right t x = If the derivtive exists ner x =, this mens f (x) > 0 when x is ner nd x <, becuse the function must slope up just º¾ Ì Ö Ø Ö Ú Ø Ú Ø Ø to the left of Similrly, f (x) < 0 when x is ner nd x >, becuse f slopes down from the locl mximum s we move to the right Using the sme resoning, if there is locl minimum t x =, the derivtive of f must be negtive just to the left of nd positive just to the right If the derivtive exists ner but does not chnge from positive to negtive or negtive to positive, tht is, it is postive on both sides or negtive on both sides, then there is neither mximum nor minimum when x = See the first grph in figure 51 nd the grph in figure 52 for exmples EXAMPLE 54 Find ll locl mximum nd minimum points for f(x) = sin x + cos x using the first derivtive test The derivtive is f (x) = cos x sin x nd from exmple 53 the criticl vlues we need to consider re π/4 nd 5π/4 The grphs of sinx nd cos x re shown in figure 54 Just to the left of π/4 the cosine is lrger thn the sine, so f (x) is positive; just to the right the cosine is smller thn the sine, so f (x) is negtive This mens there is locl mximum t π/4 Just to the left of 5π/4 the cosine is smller thn the sine, nd to the right the cosine is lrger thn the sine This mens tht the derivtive f (x) is negtive to the left nd positive to the right, so f hs locl minimum t 5π/4 Exercises 52 π 4 Figure 54 5π 4 The sine nd cosine In 1 13, find ll criticl points nd identify them s locl mximum points, locl minimum points, or neither 1 y = x 2 x 2 y = 2 + 3x x 3 3 y = x 3 9x x 4 y = x 4 2x y = 3x 4 4x 3 6 y = (x 2 1)/x 7 y = 3x 2 (1/x 2 ) 14 Find the mxim nd minim of f(x) = sec x 8 y = cos(2x) x 9 f(x) = (5 x)/(x + 2) 10 f(x) = x f(x) = x 3 /(x + 1) j 12 f(x) = x 2 sin(1/x) x 0 0 x = 0 13 f(x) = sin 2 x

4 53 The second derivtive test Chpter 5 Curve Sketching 15 Let f(θ) = cos 2 (θ) 2sin(θ) Find the intervls where f is incresing nd the intervls where f is decresing inside of [0,2π] Use this informtion to clssify the criticl points of f s either locl mximums, locl minimums, or neither 16 Let r > 0 Find the locl mxim nd minim of the function f(x) = r 2 x 2 on its domin ( r, r) Sketch the curve nd explin why the result is unsurprising 17 Let f(x) = x 2 + bx + c with 0 Show tht f hs exctly one criticl point using the first derivtive test Give conditions on nd b which gurntee tht the criticl point will be mximum It is possible to see this without using clculus t ll; explin The bsis of the first derivtive test is tht if the derivtive chnges from positive to negtive t point t which the derivtive is zero then there is locl mximum t the point, nd similrly for locl minimum If f chnges from positive to negtive it is decresing; this mens tht the derivtive of f, f, might be negtive, nd if in fct f is negtive then f is definitely decresing, so there is locl mximum t the point in question Note well tht f might chnge from positive to negtive while f is zero, in which cse f gives us no informtion bout the criticl vlue Similrly, if f chnges from negtive to positive there is locl minimum t the point, nd f is incresing If º Ì ÓÒ Ö Ú Ø Ú Ø Ø f > 0 t the point, this tells us tht f is incresing, nd so there is locl minimum EXAMPLE 55 Consider gin f(x) = sin x + cos x, with f (x) = cos x sin x nd f (x) = sin x cos x Since f (π/4) = 2/2 2/2 = 2 < 0, we know there is locl mximum t π/4 Since f (5π/4) = 2/2 2/2 = 2 > 0, there is locl minimum t 5π/4 When it works, the second derivtive test is often the esiest wy to identify locl mximum nd minimum points Sometimes the test fils, nd sometimes the second derivtive is quite difficult to evlute; in such cses we must fll bck on one of the previous tests 2 y = 2 + 3x x 3 3 y = x 3 9x x 4 y = x 4 2x y = 3x 4 4x 3 6 y = (x 2 1)/x 7 y = 3x 2 (1/x 2 ) 8 y = cos(2x) x 9 y = 4x + 1 x 10 y = (x + 1)/ p 5x y = x 5 x 12 y = 6x + sin 3x 13 y = x + 1/x 14 y = x 2 + 1/x 15 y = (x + 5) 1/4 16 y = tn 2 x 17 y = cos 2 x sin 2 x 18 y = sin 3 x We know tht the sign of the derivtive tells whether function is incresing or decresing For exmple, ny time tht f (x) > 0, tht mens tht f(x) is incresing The sign of the second derivtive f (x) tells us whether f is incresing or decresing; we hve seen tht if f is zero nd incresing t point then there is locl minimum t the point, nd if f is zero nd decresing t point then there is locl mximum t the point Thus, we extrcted informtion bout f from informtion bout f We cn get informtion from the sign of f even when f is not zero Suppose tht f () > 0 This mens tht ner x = f is incresing If f () > 0, this mens tht f º ÓÒ Ú ØÝ Ò Ò Ð Ø ÓÒÔÓ ÒØ slopes up nd is getting steeper; if f () < 0, this mens tht f slopes down nd is getting less steep The two situtions re shown in figure 55 A curve tht is shped like this is clled concve up EXAMPLE 56 Let f(x) = x 4 The derivtives re f (x) = 4x 3 nd f (x) = 12x 2 Zero is the only criticl vlue, but f (0) = 0, so the second derivtive test tells us nothing However, f(x) is positive everywhere except t zero, so clerly f(x) hs locl minimum t zero On the other hnd, f(x) = x 4 lso hs zero s its only criticl vlue, nd the second derivtive is gin zero, but x 4 hs locl mximum t zero Figure 55 f () > 0: f () positive nd incresing, f () negtive nd incresing Exercises 53 Find ll locl mximum nd minimum points by the second derivtive test Now suppose tht f () < 0 This mens tht ner x = f is decresing If f () > 0, this mens tht f slopes up nd is getting less steep; if f () < 0, this mens tht f slopes 1 y = x 2 x

5 54 Concvity nd inflection points Chpter 5 Curve Sketching Figure 56 f () < 0: f () positive nd decresing, f () negtive nd decresing down nd is getting steeper The two situtions re shown in figure 56 A curve tht is shped like this is clled concve down If we re trying to understnd the shpe of the grph of function, knowing where it is concve up nd concve down helps us to get more ccurte picture Of prticulr interest re points t which the concvity chnges from up to down or down to up; such points re clled inflection points If the concvity chnges from up to down t x =, f chnges from positive to the left of to negtive to the right of, nd usully f () = 0 We cn identify such points by first finding where f (x) is zero nd then checking to see whether f (x) does in fct go from positive to negtive or negtive to positive t these points Note tht it is possible tht f () = 0 but the concvity is the sme on both sides; f(x) = x 4 is n exmple EXAMPLE 57 Describe the concvity of f(x) = x 3 x f (x) = 3x 2 1, f (x) = 6x Since f (0) = 0, there is potentilly n inflection point t zero Since f (x) > 0 when x > 0 nd f (x) < 0 when x < 0 the concvity does chnge from down to up t zero, nd the curve is concve down for ll x < 0 nd concve up for ll x > 0 Note tht we need to compute nd nlyze the second derivtive to understnd concvity, so we my s well try to use the second derivtive test for mxim nd minim If for some reson this fils we cn then try one of the other tests Exercises 54 Describe the concvity of the functions in y = x 2 x 2 y = 2 + 3x x 3 3 y = x 3 9x x 4 y = x 4 2x y = 3x 4 4x 3 6 y = (x 2 1)/x 7 y = 3x 2 (1/x 2 ) 8 y = sin x + cosx 9 y = 4x + 1 x 10 y = (x + 1)/ p 5x y = x 5 x 12 y = 6x + sin 3x 13 y = x + 1/x 14 y = x 2 + 1/x 15 y = (x + 5) 1/4 16 y = tn 2 x 17 y = cos 2 x sin 2 x 18 y = sin 3 x 19 Identify the intervls on which the grph of the function f(x) = x 4 4x is of one of these four shpes: concve up nd incresing; concve up nd decresing; concve down nd incresing; concve down nd decresing 20 Describe the concvity of y = x 3 + bx 2 + cx + d You will need to consider different cses, depending on the vlues of the coefficients 21 Let n be n integer greter thn or equl to two, nd suppose f is polynomil of degree n How mny inflection points cn f hve? Hint: Use the second derivtive test nd the fundmentl theorem of lgebr º ÝÑÔØÓØ Ò ÇØ ÖÌ Ò ØÓÄÓÓ ÓÖ A verticl symptote is plce where the function becomes infinite, typiclly becuse the formul for the function hs denomintor tht becomes zero For exmple, the reciprocl function f(x) = 1/x hs verticl symptote t x = 0, nd the function tnx hs verticl symptote t x = π/2 (nd lso t x = π/2, x = 3π/2, etc) Whenever the formul for function contins denomintor it is worth looking for verticl symptote by checking to see if the denomintor cn ever be zero, nd then checking the limit t such points Note tht there is not lwys verticl symptote where the derivtive is zero: f(x) = (sinx)/x hs zero denomintor t x = 0, but since lim x 0 (sin x)/x = 1 there is no symptote there A horizontl symptote is horizontl line to which f(x) gets closer nd closer s x pproches (or s x pproches ) For exmple, the reciprocl function hs the x-xis for horizontl symptote Horizontl symptotes cn be identified by computing the limits lim x f(x) nd lim x f(x) Since lim x 1/x = lim x 1/x = 0, the line y = 0 (tht is, the x-xis) is horizontl symptote in both directions Some functions hve symptotes tht re neither horizontl nor verticl, but some other line Such symptotes re somewht more difficult to identify nd we will ignore them If the domin of the function does not extend out to infinity, we should lso sk wht hppens s x pproches the boundry of the domin For exmple, the function y = f(x) = 1/ r 2 x 2 hs domin r < x < r, nd y becomes infinite s x pproches either r or r In this cse we might lso identify this behvior becuse when x = ±r the denomintor of the function is zero

6 55 Asymptotes nd Other Things to Look For Chpter 5 Curve Sketching If there re ny points where the derivtive fils to exist ( cusp or corner), then we should tke specil note of wht the function does t such point Finlly, it is worthwhile to notice ny symmetry A function f(x) tht hs the sme vlue for x s for x, ie, f( x) = f(x), is clled n even function Its grph is symmetric with respect to the y-xis Some exmples of even functions re: x n when n is n even number, cos x, nd sin 2 x On the other hnd, function tht stisfies the property f( x) = f(x) is clled n odd function Its grph is symmetric with respect to the origin Some exmples of odd functions re: x n when n is n odd number, sinx, nd tn x Of course, most functions re neither even nor odd, nd do not hve ny prticulr symmetry decresing; concve down nd incresing; concve down nd decresing Discuss how the vlue of ffects these fetures Exercises 55 Sketch the curves Identify clerly ny interesting fetures, including locl mximum nd minimum points, inflection points, symptotes, nd intercepts 1 y = x 5 5x 4 + 5x 3 2 y = x 3 3x 2 9x y = (x 1) 2 (x + 3) 2/3 4 x 2 + x 2 y 2 = 2 y 2, > 0 5 y = xe x 6 y = (e x + e x )/2 7 y = e x cos x 8 y = e x sinx 9 y = e x /x 10 y = 4x + 1 x 11 y = (x + 1)/ p 5x y = x 5 x 13 y = 6x + sin3x 14 y = x + 1/x 15 y = x 2 + 1/x 16 y = (x + 5) 1/4 17 y = tn 2 x 18 y = cos 2 x sin 2 x 19 y = sin 3 x 20 y = x(x 2 + 1) 21 y = x 3 + 6x 2 + 9x 22 y = x/(x 2 9) 23 y = x 2 /(x 2 + 9) 24 y = 2 x x 25 y = 3 sin(x) sin 3 (x), for x [0, 2π] 26 y = (x 1)/(x 2 ) For ech of the following five functions, identify ny verticl nd horizontl symptotes, nd identify intervls on which the function is concve up nd incresing; concve up nd decresing; concve down nd incresing; concve down nd decresing 27 f(θ) = sec(θ) 28 f(x) = 1/(1 + x 2 ) 29 f(x) = (x 3)/(2x 2) 30 f(x) = 1/(1 x 2 ) 31 f(x) = 1 + 1/(x 2 ) 32 Let f(x) = 1/(x 2 2 ), where 0 Find ny verticl nd horizontl symptotes nd the intervls upon which the given function is concve up nd incresing; concve up nd

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