I. THE FIRST DERIVATIVE TEST: CURVE SKETCHING Let's take an arbitrary function like the one whose graph is given below: As goes from a to p, the graph rises as moves to the right towards the interval P, (a, p) and the value of f increases. We say that the function is increasing on the open interval (a, p). As continues to move towards q the graph falls and the value of f decreases, we would say that the open interval (p,q) is decreasing. Since f(r) f() for all in (a, b), we say that the function f has its maimum value on (a, b) when = r and that f(r) is the maimum value of f on interval (a, b). Notice also that f(q) f() for all in interval (a, b), we say that the function ƒ has its minimum value on interval (a, b) when = q and that f(q) is the minimum value of f on interval (a, b). Eample 1: Find the critical numbers of f if f() = 3-6 - 18 + 7 f '()= 6-1 -18 6-1 -18 = 0 find f '() and set it = 0 (6 + 6) ( - 3) = 0 factor 6 + 6 = 0 or - 3 = 0 solve for = -1 or = 3 So, the critical numbers are = -1 and = 3. Eample : Find the intervals on which ƒ is increasing if f() = 3-6 - 18 + 7 f is increasing if f ' is positive, so we need to find when f '() >0 or 6-1 - 18 > 0. From Eample 1 we know that the critical points are = -1 and = 3. We determine the sign of f ' on the intervals between critical points by using test numbers that fall to the left, between, and to the right of our critical points and plug them into f '(). (a) From the interval (-, -1) we choose = - as a test number: f ' (-) = 6(-) - 1(-) - 18 = 30 > 0 f is increasing on (-, -1) (b) From the interval (-1, 3) we choose = 0 as a test number: f ' (0) = 6(0) - 1(0) - 18 = -18 < 0 f is decreasing on (-1, 3) (c) From the interval (3, ) we choose = 4 as a test number: f ' (4) = 6(4) -1(4) - 18 = 30 > 0 f in increasing on (3, ) From (a), (b), and (c) we see that f is increasing on the intervals (-, -1) and (3, ).
Eample 3: Find the relative etreme values of f (maimum and minimum) if f() = 3-6 - 18 + 7 Make the following chart from the results of Eample : Sign of f' + - + -1 3 Notice that f ' changes sign from (+) to (-) at = -1 f has a relative maimum at = -1 The value of the relative maimum is f (-1) = 17 f ' changes sign from (-) to (+) at = 3 f has a relative minimum at = 3 The value of the relative minimum is f (3) = -47 Eample 4: Sketch the graph of f if f () = 3-6 - 18 + 7 ALWAYS start left to right when graphing. Plot the maimum and minimum points: (-1, 17) and (3, -47) f is increasing on (-, -1); the graph increases until = -1 f is decreasing on (-1, 3); the graph decreases until = 3 f is increasing again on (3, ); the graph increases until = : II. CONCAVITY AND THE SECOND DERIVATIVE TEST: A graph can be either increasing or decreasing, it could concave up, down, both or neither: The concavity of the curve has nothing to do with whether the function is increasing or decreasing. See above.
Inflection Point: A point is said to be an inflection point if the graph changes concavity at that point. The Second Derivative Test: If f '(c) = 0 (c is a critical number) and f ''(c) > 0, then f has a relative minimum at = c and is concave up f ''(c) < 0, then f has a relative maimum at = c and is concave down f ''(c) = 0, the c is an inflection point and changes concavity at = c. Eample 5: Use the second derivative test to find the relative etreme values of f if f () = 3-6 - 18 + 7 f '() = 6-1 - 18 The critical numbers are = -1 and = 3 (see Eample 1) f ''() = 1-1 Now evaluate f ''() at the critical numbers: f ''(-1) = 1(-1) -1 = -4 < 0 relative maimum at = -1 f ''(3) = 1(3) -1 = 4 > 0 relative minimum at = 3 (see Eample ) Eample 6: Use the second derivative test to find the relative etreme values and the inflection points of f if f () = - 9 + 6-3 f '()= -9 + 1-3 -9 + 1-3 = 0 find f '() and set it = 0-3(-3)(-1) = 0 factor = 3 or = 1 solve for The critical numbers are: = 1 and = 3. Sign-of-f' - + - 1 3 f is decreasing on both (-, 1) and (3, ) and increasing on (1, 3). f (1) = - and f(3) = Plot the points (1, ) and (3, ) on the graph. To discuss concavity look at f '': f ''() = 1-6 = 0 = We determine the sign of f '' by choosing points that lie to the left, between or to the right of our inflection points and plugging them into f ''(). For interval (-, ) we choose = 0; f ''(0) = 1-6(0) = 1 > 0. For interval (, ) we choose = 4; f ''(4) = 1-6(4) = -1< 0. Sign-of-f '' + - the graph is c oncave up on (-, ) and concave down on (, ) = is an inflection point; f() = - 9() + 6() - () 3 =0: = is also the -intercept; f() = 0 3
III. ASYMPTOTES: Asymptotes eist only when the function consists of fractional polynomials. We have three types of asymptotes; vertical, horizontal and oblique (slant): Vertical Asymptotes: Whatever number makes the denominator zero is a vertical asymptote: 3 5 + 6 + 5 i.e., f() =, factor 3-4 down to ( 4)( + 1), 3 4 = 4 and = -1 are vertical asymptotes. Horizontal Asymptotes Look at the highest power of the numerator (call it t for top) and the highest power of the denominator (call it b for bottom); three cases are considered. Case 1: b > t The line y = 0 is a horizontal asymptote. Case.: b = t Divided the coefficient of the t term by the coefficient of the b term; the result is the horizontal asymptote. Case 3: b + 1 = t Divide the numerator by the denominator and the quotient is called an oblique (slant) asymptote. Otherwise: No asymptotes eist. Eample 7: Find the relative etreme values, the inflection points of f and sketch the graph of: 6 f () = f '() = 1 = 0 3 find f '() and set it = 0-1 = 0 set the numerator only = 0 = 1 solve for The critical numbers are: = 1 and = 0 (denominator cannot = 0, so that value is considered a critical number) 4
f is increasing on both (-, 0) and (1, ) and decreasing on (0, 1). f(0) = undefined and f(1) = -6/144 the point (1, -6/144) is a relative minimum. Plot the point (1, -6/144) on the graph. To discuss concavity look at f '': f ''()= + 36 = 0 4 = 18 and again = 0 are critical numbers Sign-of-f '' + + - 0 18 the graph is concave up on (-, 0) and (0, 18) and concave down on (18, ) = 18 is the only inflection point: Vertical asymptotes: = 0 (denominator = 0) Horizontal asymptotes: y = 0 (second case) and finally the graph: Revised: Spring 005 Created in 1994 by Ziad Diab STUDENT LEARNING ASSISTANCE CENTER (SLAC) Teas State University-San Marcos 5