# 106 Chapter 5 Curve Sketching. If f(x) has a local extremum at x = a and. THEOREM Fermat s Theorem f is differentiable at a, then f (a) = 0.

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1 5 Curve Sketching Whether we are interested in a function as a purely mathematical object or in connection with some application to the real world, it is often useful to know what the graph of the function looks like We can obtain a good picture of the graph using certain crucial information provided by derivatives of the function and certain limits º½ Å Ü Ñ Ò Å Ò Ñ A local maximum point on a function is a point (x,y) on the graph of the function whose y coordinate is larger than all other y coordinates on the graph at points close to (x,y) More precisely, (x,f(x)) is a local maximum if there is an interval (a,b) with a < x < b and f(x) f(z) for every z in (a,b) Similarly, (x,y) is a local minimum point if it has locally the smallest y coordinate Again being more precise: (x,f(x)) is a local minimum if there is an interval (a,b) with a < x < b and f(x) f(z) for every z in (a,b) A local extremum is either a local minimum or a local maximum Local maximum and minimum points are quite distinctive on the graph of a function, and are therefore useful in understanding the shape of the graph In many applied problems wewanttofindthelargestorsmallestvaluethatafunctionachieves(forexample, wemight want to find the minimum cost at which some task can be performed) and so identifying maximum and minimum points will be useful for applied problems as well Some examples of local maximum and minimum points are shown in figure 511 If (x,f(x)) is a point where f(x) reaches a local maximum or minimum, and if the derivative of f exists at x, then the graph has a tangent line and the tangent line must be horizontal This is important enough to state as a theorem, though we will not prove it 105

2 106 Chapter 5 Curve Sketching A A A B B Figure 511 Some local maximum points (A) and minimum points (B) THEOREM 511 Fermat s Theorem f is differentiable at a, then f (a) = 0 If f(x) has a local extremum at x = a and Thus, the only points at which a function can have a local maximum or minimum are points at which the derivative is zero, as in the left hand graph in figure 511, or the derivative isundefined, as in the right hand graph Any value of x for which f (x) is zero or undefined is called a critical value for f When looking for local maximum and minimum points, you are likely to make two sorts of mistakes: You may forget that a maximum or minimum can occur where the derivative does not exist, and so forget to check whether the derivative exists everywhere You might also assume that any place that the derivative is zero is a local maximum or minimum point, but this is not true A portion of the graph of f(x) = x 3 is shown in figure 512 The derivative of f is f (x) = 3x 2, and f (0) = 0, but there is neither a maximum nor minimum at (0,0) Figure 512 No maximum or minimum even though the derivative is zero Since the derivative is zero or undefined at both local maximum and local minimum points, we need a way to determine which, if either, actually occurs The most elementary approach, but one that is often tedious or difficult, is to test directly whether the y coordinates near the potential maximum or minimum are above or below the y coordinate

3 51 Maxima and Minima 107 at the point of interest Of course, there are too many points near the point to test, but a little thought shows we need only test two provided we know that f is continuous (recall that this means that the graph of f has no jumps or gaps) Suppose, for example, that we have identified three points at which f is zero or nonexistent: (x 1,y 1 ), (x 2,y 2 ), (x 3,y 3 ), and x 1 < x 2 < x 3 (see figure 513) Suppose that we compute the value of f(a) for x 1 < a < x 2, and that f(a) < f(x 2 ) What can we say about the graph between a and x 2? Could there be a point (b,f(b)), a < b < x 2 with f(b) > f(x 2 )? No: if there were, the graph would go up from (a,f(a)) to (b,f(b)) then down to (x 2,f(x 2 )) and somewhere in between would have a local maximum point (This is not obvious; it is a result of the Extreme Value Theorem, theorem 612) But at that local maximum point the derivative of f would be zero or nonexistent, yet we already know that the derivative is zero or nonexistent only at x 1, x 2, and x 3 The upshot is that one computation tells us that (x 2,f(x 2 )) has the largest y coordinate of any point on the graph near x 2 and to the left of x 2 We can perform the same test on the right If we find that on both sides of x 2 the values are smaller, then there must be a local maximum at (x 2,f(x 2 )); if we find that on both sides of x 2 the values are larger, then there must be a local minimum at (x 2,f(x 2 )); if we find one of each, then there is neither a local maximum or minimum at x 2 x 1 a b x 2 x 3 Figure 513 Testing for a maximum or minimum It is not always easy to compute the value of a function at a particular point The task is made easier by the availability of calculators and computers, but they have their own drawbacks they do not always allow us to distinguish between values that are very close together Nevertheless, because this method is conceptually simple and sometimes easy to perform, you should always consider it EXAMPLE 512 Find all local maximum and minimum points for the function f(x) = x 3 x The derivative is f (x) = 3x 2 1 This is defined everywhere and is zero at x = ± 3/3 Looking first at x = 3/3, we see that f( 3/3) = 2 3/9 Now we test two points on either side of x = 3/3, making sure that neither is farther away than the nearest critical value; since 3 < 3, 3/3 < 1 and we can use x = 0 and x = 1 Since f(0) = 0 > 2 3/9 and f(1) = 0 > 2 3/9, there must be a local minimum at

4 108 Chapter 5 Curve Sketching x = 3/3 For x = 3/3, we see that f( 3/3) = 2 3/9 This time we can use x = 0 and x = 1, and we find that f( 1) = f(0) = 0 < 2 3/9, so there must be a local maximum at x = 3/3 Of course this example is made very simple by our choice of points to test, namely x = 1, 0, 1 We could have used other values, say 5/4, 1/3, and 3/4, but this would have made the calculations considerably more tedious EXAMPLE 513 Find all local maximum and minimum points for f(x) = sinx+cosx The derivative is f (x) = cosx sinx This is always defined and is zero whenever cosx = sinx Recalling that the cosx and sinx are the x and y coordinates of points on a unit circle, we see that cosx = sinx when x is π/4, π/4±π, π/4±2π, π/4±3π, etc Since both sine and cosine have a period of 2π, we need only determine the status of x = π/4 and x = 5π/4 We can use 0 and π/2 to test the critical value x = π/4 We find that f(π/4) = 2, f(0) = 1 < 2 and f(π/2) = 1, so there is a local maximum when x = π/4 and also when x = π/4±2π, π/4±4π, etc We can summarize this more neatly by saying that there are local maxima at π/4±2kπ for every integer k We use π and 2π to test the critical value x = 5π/4 The relevant values are f(5π/4) = 2, f(π) = 1 > 2, f(2π) = 1 > 2, so there is a local minimum at x = 5π/4, 5π/4±2π, 5π/4±4π, etc More succinctly, there are local minima at 5π/4±2kπ for every integer k Exercises 51 In problems1 12, find all localmaximumandminimumpoints (x,y) bythe methodofthissection 1 y = x 2 x 2 y = 2+3x x 3 3 y = x 3 9x 2 +24x 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 { x 3 x < 3 x 1 x < 2 9 f(x) = x 2 10 f(x) = x 3 3 x 5 x 2 1/x x > 5 { 2 x = 0 11 f(x) = x 2 98x+4 12 f(x) = 1/x 2 x 0 13 For any real number x there is a unique integer n such that n x < n+1, and the greatest integer function is defined as x = n Where are the critical values of the greatest integer function? Which are local maxima and which are local minima? 14 Explain why the function f(x) = 1/x has no local maxima or minima 15 How many critical points can a quadratic polynomial function have?

5 52 The first derivative test Show that a cubic polynomial can have at most two critical points Give examples to show that a cubic polynomial can have zero, one, or two critical points 17 Explore the family of functions f(x) = x 3 + cx + 1 where c is a constant How many and what types of local extremes are there? Your answer should depend on the value of c, that is, different values of c will give different answers 18 We generalize the preceding two questions Let n be a positive integer and let f be a polynomial of degree n How many critical points can f have? (Hint: Recall the Fundamental Theorem of Algebra, which says that a polynomial of degree n has at most n roots) º¾ Ì Ö Ø Ö Ú Ø Ú Ø Ø The method of the previous section for deciding whether there is a local maximum or minimum at a critical value is not always convenient We can instead use information about the derivative f (x) to decide; since we have already had to compute the derivative to find the critical values, there is often relatively little extra work involved in this method How can the derivative tell us whether there is a maximum, minimum, or neither at a point? Suppose that f (a) = 0 If there is a local maximum when x = a, the function must be lower near x = a than it is right at x = a If the derivative exists near x = a, this means f (x) > 0 when x is near a and x < a, because the function must slope up just to the left of a Similarly, f (x) < 0 when x is near a and x > a, because f slopes down from the local maximum as we move to the right Using the same reasoning, if there is a local minimum at x = a, the derivative of f must be negative just to the left of a and positive just to the right If the derivative exists near a but does not change from positive to negative or negative to positive, that is, it is positive on both sides or negative on both sides, then there is neither a maximum nor minimum when x = a See the first graph in figure 511 and the graph in figure 512 for examples EXAMPLE 521 Find all local maximum and minimum points for f(x) = sinx+cosx using the first derivativetest The derivativeisf (x) = cosx sinxand from example513 the critical values we need to consider are π/4 and 5π/4 The graphs of sinx and cosx are shown in figure 521 Just to the left of π/4 the cosine is larger than the sine, so f (x) is positive; just to the right the cosine is smaller than the sine, so f (x) is negative This means there is a local maximum at π/4 Just to the left of 5π/4 the cosine is smaller than the sine, and to the right the cosine is larger than the sine This means that the derivative f (x) is negative to the left and positive to the right, so f has a local minimum at 5π/4 Exercises 52 In 1 13, find all critical points and identify them as local maximum points, local minimum points, or neither

6 110 Chapter 5 Curve Sketching π 4 5π 4 Figure 521 The sine and cosine 1 y = x 2 x 2 y = 2+3x x 3 3 y = x 3 9x 2 +24x 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 f(x) = (5 x)/(x+2) 10 f(x) = x { 11 f(x) = x 3 /(x+1) 12 f(x) = x 2 sin(1/x) x 0 0 x = 0 13 f(x) = sin 2 x 14 Find the maxima and minima of f(x) = secx 15 Let f(θ) = cos 2 (θ) 2sin(θ) Find the intervals where f is increasing and the intervals where f is decreasing in [0, 2π] Use this information to classify the critical points of f as either local maximums, local minimums, or neither 16 Let r > 0 Find the local maxima and minima of the function f(x) = r 2 x 2 on its domain [ r, r] 17 Let f(x) = ax 2 + bx + c with a 0 Show that f has exactly one critical point Give conditions on a and b which guarantee that the critical point will be a maximum It is possible to see this without using calculus at all; explain º Ì ÓÒ Ö Ú Ø Ú Ø Ø The basis of the first derivative test is that if the derivative changes from positive to negative at a point at which the derivative is zero then there is a local maximum at the point, and similarly for a local minimum If f changes from positive to negative it is decreasing; this means that the derivative of f, f, might be negative, and if in fact f is negative then f is definitely decreasing, so there is a local maximum at the point in question Note well that f might change from positive to negative while f is zero, in which case f gives us no information about the critical value Similarly, if f changes from negative to positive there is a local minimum at the point, and f is increasing If f > 0 at the point, this tells us that f is increasing, and so there is a local minimum

7 54 Concavity and inflection points 111 EXAMPLE 531 Consider again f(x) = sinx+cosx, with f (x) = cosx sinx and f (x) = sinx cosx Since f (π/4) = 2/2 2/2 = 2 < 0, we know there is a local maximum at π/4 Since f (5π/4) = 2/2 2/2 = 2 > 0, there is a local minimum at 5π/4 When it works, the second derivative test is often the easiest way to identify local maximum and minimum points Sometimes the test fails, and sometimes the second derivative is quite difficult to evaluate; in such cases we must fall back on one of the previous tests EXAMPLE 532 Let f(x) = x 4 The derivatives are f (x) = 4x 3 and f (x) = 12x 2 Zero is the only critical value, but f (0) = 0, so the second derivative test tells us nothing However, f(x) is positive everywhere except at zero, so clearly f(x) has a local minimum at zero On the other hand, f(x) = x 4 also has zero as its only critical value, and the second derivative is again zero, but x 4 has a local maximum at zero Exercises 53 Find all local maximum and minimum points by the second derivative test 1 y = x 2 x 2 y = 2+3x x 3 3 y = x 3 9x 2 +24x 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)/ 5x y = x 5 x 12 y = 6x+sin3x 13 y = x+1/x 14 y = x 2 +1/x 15 y = (x+5) 1/4 16 y = tan 2 x 17 y = cos 2 x sin 2 x 18 y = sin 3 x º ÓÒ Ú ØÝ Ò Ò Ð Ø ÓÒ ÔÓ ÒØ We know that the sign of the derivative tells us whether a function is increasing or decreasing; for example, when f (x) > 0, f(x) is increasing The sign of the second derivative f (x) tells us whether f is increasing or decreasing; we have seen that if f is zero and increasing at a point then there is a local minimum at the point, and if f is zero and decreasing at a point then there is a local maximum at the point Thus, we extracted information about f from information about f We can get information from the sign of f even when f is not zero Suppose that f (a) > 0 This means that near x = a, f is increasing If f (a) > 0, this means that f slopes up and is getting steeper; if f (a) < 0, this means that f slopes down and is getting

8 112 Chapter 5 Curve Sketching a a Figure 541 f (a) > 0: f (a) positive and increasing, f (a) negative and increasing less steep The two situations are shown in figure 541 A curve that is shaped like this is called concave up Now suppose that f (a) < 0 This means that near x = a, f is decreasing If f (a) > 0, this means that f slopes up and is getting less steep; if f (a) < 0, this means that f slopes down and is getting steeper The two situations are shown in figure 542 A curve that is shaped like this is called concave down a a Figure 542 f (a) < 0: f (a) positive and decreasing, f (a) negative and decreasing If we are trying to understand the shape of the graph of a function, knowing where it is concave up and concave down helps us to get a more accurate picture Of particular interest are points at which the concavity changes from up to down or down to up; such points are called inflection points If the concavity changes from up to down at x = a, f changes from positive to the left of a to negative to the right of a, and usually f (a) = 0 We can identify such points by first finding where f (x) is zero and then checking to see whether f (x) does in fact go from positive to negative or negative to positive at these points Note that it is possible that f (a) = 0 but the concavity is the same on both sides; f(x) = x 4 at x = 0 is an example EXAMPLE 541 Describe the concavity of f(x) = x 3 x f (x) = 3x 2 1, f (x) = 6x Since f (0) = 0, there is potentially an inflection point at zero Since f (x) > 0 when x > 0 and f (x) < 0 when x < 0 the concavity does change from down to up at zero, and the curve is concave down for all x < 0 and concave up for all x > 0 Note that we need to compute and analyze the second derivative to understand concavity, so we may as well try to use the second derivative test for maxima and minima If for some reason this fails we can then try one of the other tests

9 55 Asymptotes and Other Things to Look For 113 Exercises 54 Describe the concavity of the functions in y = x 2 x 2 y = 2+3x x 3 3 y = x 3 9x 2 +24x 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 = sinx+cosx 9 y = 4x+ 1 x 10 y = (x+1)/ 5x y = x 5 x 12 y = 6x+sin3x 13 y = x+1/x 14 y = x 2 +1/x 15 y = (x+5) 1/4 16 y = tan 2 x 17 y = cos 2 x sin 2 x 18 y = sin 3 x 19 Identify the intervals on which the graph of the function f(x) = x 4 4x is of one of these four shapes: concave up and increasing; concave up and decreasing; concave down and increasing; concave down and decreasing 20 Describe the concavity of y = x 3 +bx 2 + cx +d You will need to consider different cases, depending on the values of the coefficients 21 Let n be an integer greater than or equal to two, and suppose f is a polynomial of degree n How many inflection points can f have? Hint: Use the second derivative test and the fundamental theorem of algebra º ÝÑÔØÓØ Ò ÇØ Ö Ì Ò ØÓ ÄÓÓ ÓÖ A vertical asymptote is a place where the function becomes infinite, typically because the formula for the function has a denominator that becomes zero For example, the reciprocal function f(x) = 1/x has a vertical asymptote at x = 0, and the function tanx has a vertical asymptote at x = π/2 (and also at x = π/2, x = 3π/2, etc) Whenever the formula for a function contains a denominator it is worth looking for a vertical asymptote by checking to see if the denominator can ever be zero, and then checking the limit at such points Note that there is not always a vertical asymptote where the denominator is zero: f(x) = (sinx)/x has a zero denominator at x = 0, but since lim(sinx)/x = 1 there is no x 0 asymptote there A horizontal asymptote is a horizontal line to which f(x) gets closer and closer as x approaches (or as x approaches ) For example, the reciprocal function has the x-axis for a horizontal asymptote Horizontal asymptotes can be identified by computing the limits lim f(x) and lim f(x) Since lim 1/x = lim 1/x = 0, the line y = 0 (that x x x x is, the x-axis) is a horizontal asymptote in both directions

10 114 Chapter 5 Curve Sketching Some functions have asymptotes that are neither horizontal nor vertical, but some other line Such asymptotes are somewhat more difficult to identify and we will ignore them If the domain of the function does not extend out to infinity, we should also ask what happens as x approaches the boundary of the domain For example, the function y = f(x) = 1/ r 2 x 2 has domain r < x < r, and y becomes infinite as x approaches either r or r In this case we might also identify this behavior because when x = ±r the denominator of the function is zero If there are any points where the derivative fails to exist (a cusp or corner), then we should take special note of what the function does at such a point Finally, it is worthwhile to notice any symmetry A function f(x) that has the same value for x as for x, ie, f( x) = f(x), is called an even function Its graph is symmetric with respect to the y-axis Some examples of even functions are: x n when n is an even number, cosx, and sin 2 x On the other hand, a function that satisfies the property f( x) = f(x) is called an odd function Its graph is symmetric with respect to the origin Some examples of odd functions are: x n when n is an odd number, sinx, and tanx Of course, most functions are neither even nor odd, and do not have any particular symmetry Exercises 55 Sketch the curves Identify clearly any interesting features, including local maximum and minimum points, inflection points, asymptotes, and intercepts 1 y = x 5 5x 4 +5x 3 2 y = x 3 3x 2 9x+5 3 y = (x 1) 2 (x+3) 2/3 4 x 2 +x 2 y 2 = a 2 y 2, a > 0 5 y = xe x 6 y = (e x +e x )/2 7 y = e x cosx 8 y = e x sinx 9 y = e x /x 10 y = 4x+ 1 x 11 y = (x+1)/ 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 = tan 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 = 3sin(x) sin 3 (x), for x [0,2π] 26 y = (x 1)/(x 2 ) For each of the following five functions, identify any vertical and horizontal asymptotes, and identify intervals on which the function is concave up and increasing; concave up and decreasing; concave down and increasing; concave down and decreasing

11 55 Asymptotes and Other Things to Look For 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 a 2 ), where a 0 Find any vertical and horizontal asymptotes and the intervals upon which the given function is concave up and increasing; concave up and decreasing; concave down and increasing; concave down and decreasing Discuss how the value of a affects these features

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