Answers for Calculus Review 4.1-4.4 (Extrema and Concavity) 1. A critical number is a value of the independent variable (a/k/a x) in the domain of the function at which the derivative is zero or undefined. The title of the quiz is Extrema and Concavity, and the significance of critical numbers in that respect is that extrema of a function can only occur at critical numbers or at endpoints of the domain. 2. Absolute extrema are the largest and smallest values that a function takes on in its domain. Local extrema don t have to be the biggest/smallest everywhere, just in their immediate vicinity. (I m the oldest person who regularly teaches in my classroom, but not the oldest person who regularly teaches in the school. So I m a local maximum.) 3. Absolute extrema are also called global extrema (because they consider the total behavior of the function), while local extrema are also called relative extrema, because they are only the extreme values relative to their immediate vicinities. 4. A function f is decreasing on an interval when, for any two values x 1 and x 2 on the interval, if x 1 < x 2, then f (x 1 ) > f (x 2 ). (Get it? As x gets bigger, y gets smaller, so the function decreases!) 5. A point of inflection occurs at a point on the graph where the sign of the second derivative changes. Note that it is not enough for the second derivative to be zero. 6. f (x) = 3x 5 5x 3 1; [ 2, 2] a) Find the critical numbers. Critical numbers are values of the variable at which the derivative of the function is either zero or undefined. So first I ll need a derivative. f (x) = 15x 4 15x 2 Since f is a polynomial, it is always defined; I only need to determine when f is zero. (Without a calculator!!!) f = 0 15x 4 15x 2 = 0. Factor. 15x 2 (x 2 1) = 15x 2 (x 1)(x + 1) = 0 So x = 0, 1, 1 are the critical numbers. b) The easiest way to check if a function is increasing or decreasing is to examine the sign of the derivative. The function is a polynomial, so there are no places where it s not defined; I just need to look at the signs of the derivative in the intervals determined by the critical numbers. I like number lines for this. The endpoints of the segment are 2 and 2 because I was given a finite interval. To get the signs, I just looked at what would be true of the derivative in each interval. Here s a more detailed analysis. Remember that f (x) = 15x 2 (x 1)(x + 1). Between 2 and 1, I ll use x = 1.5. So f ( 1.5) = 15( 1.5) 2 ( 1.5 1)( 1.5 + 1) = 15 positive negative negative = positive. That s where the first plus sign over the number line came from. I did the same thing with a number in each of the other intervals (which I will not bore either you or me with here). The little line segments under each interval are to remind me that a + on the first derivative means that f is increasing, and a means it s decreasing there. f is increasing on [ 2, 1] and on [1, 2] because f is positive there, and decreasing on [ 1, 1], where f is negative. (I ll explain more about the open versus closed intervals after question 10, when I can refer back to all of the questions like this.) c) To find whether each critical number is the location of a maximum, minimum, or neither, I ll use the first derivative test. It says that, for a continuous function at a critical point, if the sign of the first derivative changes from positive to negative, there s a local maximum; if it changes from negative to positive, there s a local minimum; and if the sign does not p. 1 of 6
change, there is no local extreme value at that location. The three critical numbers were 1, 0, and 1. At x = 1, the sign of f (x) changed from positive to negative, producing a local maximum. At x = 0, the sign of f (x) did not change, indicating that there is no local extremum there. And at x = 1, the sign of f (x) changed from negative to positive, showing a local minimum. d) To find the values of the absolute extrema, I ll check the value of f at the location of each local extreme and at both endpoints. x 2 1 1 2 f(x) 57 1 3 55 Clearly, the absolute minimum value is 57 and the absolute maximum value is 55. In this case, none of the local extrema happens to be an absolute extreme. Notice that the question asks for values rather than points, so the answers are y-coordinates. e) Points of inflection are located where the sign of the second derivative changes. To find those locations, first determine where the second derivative is zero or undefined. f (x) = 15x 4 15x 2 f (x) = 60x 3 30x As a polynomial, f (x) is never undefined. f (x) = 60x 3 30x = 0 30x(2x 2 1) = 0 30x = 0 or 2x 2 1 = 0 1 x 0 or x 2 Now that we know where the boundaries are, a number line makes sense. Points of inflection are located at 1 1 x =, 0, and, since that s 2 2 where f changes signs. But I still have to find y-values, since points of inflection are, you know, points. Before you panic, I would not ask you to plug in anything quite as ugly as this without a calculator. (And even if I did, you wouldn t have to simplify it.) 5 3 f 1 1 1 3 5 1 2 2 2 3 5 7 1 1 4 2 2 2 4 2 f (0) = 3(0) 5 5(0) 3 1 = 1 5 3 f 1 1 1 3 5 1 2 2 2 3 5 7 1 1 4 2 2 2 4 2 And finally, the points of inflection are 1 7, 1, (0, 1), and 2 4 2 1 7, 1 2 4 2. 7. g(x) = 3 + sin x; [0, 2π] a) Find the critical numbers. g(x) = cos x g is never undefined. g = 0 when x =, (and lots of other places, but these are the only ones on the specified interval) Therefore the only critical numbers for g are and. b) To determine where g is increasing and where it s decreasing, I ll look at the signs of the derivative. Here s a number line. And I can conclude that g is increasing on 0, 2 and 3,2 2 because g is 3 positive there, and decreasing on, 2 2, where g is negative. c) Using the first derivative test, g has a relative maximum at x = 2 because g p. 2 of 6
changes signs from positive to negative there, and a relative minimum at x = 3 because g changes from 2 negative to positive there. d) Absolute extrema can occur at critical numbers or at endpoints. I can just compare the values of g at each of those numbers to find the largest and smallest. Here s a table. x 0 2π g(x) 3 4 2 3 Based on these values, the local maximum of 4 is also the absolute maximum and the local minimum of 2 is also the absolute minimum. e) Points of inflection occur where the second derivative changes signs. Since g(x) = cos x, I differentiate again to find that g(x) = sin x. The second derivative has zeroes at x = 0,, and 2 on the given domain, and only changes signs at x =. (Think about a graph of g if this isn t obvious to you.) So the point of inflection has a y- coordinate of g() = 3 + sin = 3. The only point of inflection is (, 3). 2 x, 1 x1 8. hx ( ) 2 2 x 1 x 3 a) To find the critical numbers, I need the derivative. The original function is piecewise, so the derivative is piecewise, too. 2, 1 x 1 h( x) 2 x, 1 x3 The derivative is undefined at x = 1, since the derivative is not the same as x approaches 1 from the left and the right. Thus there can t be a or sign at x = 1. 1 If h = 0, that must mean that x = 0 on the second part of the function. However, 0 isn t in the 1 The one-sided derivatives exist at the endpoints, so I have included them, but it would have been okay to use < there, too. domain of the second part of the function, so the slope is never 0. That means the only critical number is x = 1, where the derivative is undefined. b) For increasing and decreasing, I ll look at the signs of the derivative. There are only two relevant intervals this time. h is increasing on [ 1, 1) where h > 0, and decreasing on [1, 3] where h < 0. c) Using the first derivative test is not actually possible this time! That test requires that the function is continuous. But if you look at the definition of h, lim h x 2 lim h x 1. As x = 1 is x1 and x1 part of the right-hand side, this means that the point is neither a maximum nor a minimum. A graph should help you see why. d) The only critical point is neither a maximum nor a minimum. At the left endpoint, h( 1) = 2, and at the right endpoint, h(3) = 7. The absolute minimum is 7 and there is no absolute maximum. e) Points of inflection are a little weird here. The left part of the graph has no concavity at all; the right half is only concave down. The second derivative looks like this: 0, 1 x 1 h x. 2, 1 x 3 The sign of the second derivative doesn t really change at x = 1, since 0 is neither positive nor negative. I would say that there are no points of inflection. Okay, I promised some commentary about the open and closed intervals in the previous four questions. To simplify what comes next, I ll just write about increasing behavior. The same argument applies equally well to decreasing behavior. Remember that the definition of increasing doesn t depend on derivatives. If, as x-values increase, the y- values increase, then the function is increasing. Period. p. 3 of 6
So why does that matter? The way that we re testing for increasing behavior involves the derivative, and the boundary values we re using are critical numbers: places where, by definition, the derivative is neither positive nor negative. So in the number lines, I was really using open intervals. It wouldn t be true to say that the derivative is positive on a closed interval if, in fact, it s zero at the endpoints of the interval. But how, then, am I able to say that the function is increasing on a closed interval? Because, since the continuous function is behaving that way all the way right up to the very edge of the interval and the function still exists there and the derivative hasn t changed signs, that last y-value on the end must still meet the definition of increasing; it must be greater than the y-value just to its left. I hope this helps. Just keep in mind that, from the top of a hill, every direction is down increasing towards the top, decreasing from the top. For the next set of problems, I ve just drawn one possible graph for each situation. There are literally infinitely many possibilities for each one, but these will do. There s not a lot to say about some of them. 9. The function never changes concavity, has a local maximum at x = 2, and is differentiable at x = 2. If it s differentiable at a local maximum, the derivative must be zero there. Yes, this is also an absolute maximum. 10. The function has a relative maximum at x = 2 and is continuous but not differentiable at x = 2. To be continuous but not differentiable means a cusp, a corner, or a vertical tangent. Of those, a vertical tangent wouldn t work, because the function wouldn t have a maximum there. 11. The function has a relative maximum at x = 2 and is discontinuous at x = 2. The filled circle has to be the higher of the two if there s to be a maximum. 12. The function has exactly two local maximum points, exactly one local minimum point, and no absolute minimum. The fact that there s no absolute minimum means that this must either go to or have open circles on the ends. I didn t draw the arrows, but the lack of specified endpoints implies that it continues in the same fashion. 13. The function is always increasing, has two vertical asymptotes, and exactly one point of inflection. As I ve drawn it, there is also a horizontal asymptote at y = 0. The function is increasing on its entire domain, but it s not always continuous. 14. The function has domain [ 3, 2] and exactly one global extreme, which occurs at infinitely many x-values. A continuous function on a closed interval must have both an absolute maximum and an absolute minimum. Because the question specifies only one global extreme, they must be the same. I know it s a closed interval. 15. The function has an absolute minimum at a point of discontinuity. p. 4 of 6
It doesn t have to be a separate dot; the graph from #11 flipped upside down would have done just as well. 16. The function has exactly 3 critical numbers, but no absolute extrema. Three critical numbers means that there can be only three places where the derivative is zero or nonexistent. I chose to use horizontal tangents at all three of them, but I could have used places where the derivative didn t exist. The ends go to and, so there are no absolute extrema. 17. The function has exactly 3 critical numbers, but no local or absolute extrema. Pretty? No. I made it ugly on purpose. I could have gone with multiple places like the origin in #16. I just wanted you to see a different type. 18. The function has exactly one relative maximum point, no relative minima, and exactly two critical points. No relative minima implies either open circles or negative infinity at the ends. I went with decreasing forever. 19. The Extreme Value Theorem says that if f is continuous on the closed interval [a, b], then it must have both a maximum and a minimum value on that interval. This means that a continuous function will have a y- value that s bigger than (or tied for biggest with) all the other y-values and a y-value that s smaller than (or tied for smallest with) all the other y-values. It would apply to problems 6 and 7 on this worksheet, but not to problem 8. 20. Rolle s theorem says that if f is continuous on [a, b] and differentiable on (a, b) and if f (a) = f (b), then there must exist a value of c between a and b at which f (c) = 0. This means that for a differentiable function with both endpoints at the same height, a critical number is guaranteed somewhere on the function. 21. The Mean Value Theorem says that if f is continuous on [a, b] and differentiable on (a, b), there exists a value of c between a and b at f b f a which fc. This means that, b a for a differentiable function on a closed interval, there will be a point at which the slope of the tangent line is the same as the slope of the secant line between its endpoints. Equivalently, this means that there s a place at which the instantaneous rate of change is equal to the average rate of change. 22. (a) Extrema can only happen at critical numbers and at endpoints. Critical numbers are locations in the interior of the domain where f (x) is either zero or nonexistent. In this problem, those would be x = 1 and x = 2. (The reason we do not include x = 0 is that it s an endpoint, but we ll be checking that point, too.) To decide if there is a maximum or minimum at each location, a number line can help. I labeled the critical numbers and endpoints, and put in the signs of f (which is also labeled). Since f changes signs from negative to positive at x = 2, that must be the location of a local minimum. Since f is negative to the right of x = 0, that must be the location of an endpoint maximum. (Think about it; the graph is decreasing as you move away from the endpoint, so it must have started out up high.) And since f is positive to the left of x = 3, that must be an endpoint maximum as well. But how do we decide about absolute extrema. There s only one possible location of the absolute minimum: it must be at x = 2, where the local minimum is. p. 5 of 6
But is f (0) or f (3) the absolute maximum? The absolute max has to occur at x = 0, because f (0) = 1, and just to the left of x = 3, the values of f are negative. We know f is continuous, so the biggest possible value of f is 0 there. To summarize: absolute maximum at x = 0, absolute minimum at x = 2. (b) Points of inflection happen where f changes signs. You can see in the table that f changes from positive to negative at x = 1; that s it. (c) Here s a graph. I started with the points (0, 1), (1, 0), and (2, 1). p. 6 of 6