Curve Sketching with Derivatives

Curve Sketching The first derivative of a function tells us over which intervals the function is increasing and decreasing, and allows us find the extrema of a function, which gives us some idea of how a function looks. However, knowing just the sign of the first derivative over an interval does not tell us exactly how a function looks. We also need to know something about the magnitude of the first derivative, but we can’t simply evaluate the first derivative at all points within an interval, as there are too many points to do so. We’ll have to use a different method. Consider the function x3 . 4 3 2 x3 1 0 -1 -2 -3 -4 -2 -1.5 -1 -0.5 0 x 0.5 1 1.5 2 Looking at the first derivative, 3x2 , we see that this function is increasing at all points other than x = 0 (because 3x2 > 0 for all x 6= 0). As x → 0− , we see that the function begins to increase more slowly, corresponding to a decreasing first derivative, and as x becomes larger than 0, the rate at which the function increases increases, corresponding to an increasing first derivative. The changing magnitude of the first derivative lends curvature to the function (compare this to a straight line, which has no curvature, as the the first derivative never changes). We say a function is concave up on an interval if the first derivative is increasing on that interval, which corresponds to a positive second derivative over the interval. Similarly, we say a function is concave down on an interval if the first derivative is decreasing on that interval, which corresponds to a negative second derivative over the interval. Thus, the function x3 is concave up for x > 0, and concave down for x < 0. Finally, note that at x = 0, the concavity of the function x3 changes; we call such a point a point of an inflection. A point of inflection is any point where the first derivative exists, and the concavity of the function changes around that point. The information that the second derivative tells us about concavity is summarized below. 1. The second derivative is positive (f 00 (x) > 0): When the second derivative is positive, the function f (x) is concave up. 2. The second derivative is negative (f 00 (x) < 0): When the second derivative is negative, the function f (x) is concave down. 3. The second derivative is zero (f 00 (x) = 0): When the second derivative is zero, it corresponds to a possible inflection point. If the second derivative changes sign around the zero (from positive to negative, or negative to positive), then the point is an inflection point. This corresponds to a point where the function f (x) changes concavity. If the second derivative does not change sign (ie. it goes from positive to zero to positive), then it is not an inflection point (x = 0 with f (x) = x4 is an example of this). If the first derivative of a function is 0 at a point c, then we can often use the second derivative to classify the critical point c. Suppose the second derivative is positive on an interval around the critical point c, which implies that the first derivative is continuous and increasing on this interval. It follows that the first derivative must begin negative in order to increase to 0, and become positive after increasing beyond 0. In other words, such a point is a local minimum. There is a similar result for a negative second derivative, which we summarize in the following theorem. Second Derivative Test for Local Extrema Suppose f 0 (c) = 0, and f 00 is continuous on an open interval around x = c. 1. If f 00 (c) < 0, then f has a local maximum at x = c 2. If f 00 (c) > 0, then f has a local minimum at x = c 3. If f 00 (c) = 0, then the test is inconclusive. f may have a local maximum, minimum, or neither at x = c. Why do we require that f 00 be continuous in the hypothesis of the above theorem? The reason is as follows. If f 00 is continuous, then the fact that f 00 (c) > 0 or f 00 (c) < 0 implies that f 00 > 0 or f 00 < 0 on some interval around x = c, because the function is continuous so it can’t jump. It’s noteworthy that in many cases we cannot apply this theorem. First, we cannot use it for critical points at which the first derivative does not exist. Second, it does not give us any information for when the second derivative is 0 as well (such as x3 , x4 , etc). Finally, it is not useful for classifying endpoints. Despite these restrictions it is useful, for the simple fact that when we want to graph a function, we need to find the second derivative anyway. Now that we know how to interpret the first and second derivatives, let’s apply this information for curve sketching. Example 1 Sketch f (x) = x3 on [−2, 2] using the first and second derivatives. Solution We are already familiar with the graph of this function, so this will be a bit of a warmup. By repeated applications of the power rule, we find that f 0 (x) = 3x2 , and f 00 (x) = 6x. f 0 (0) = 0 and f 00 (0) = 0, so x = 0 is a critical point and a possible inflection point, but we cannot use the second derivative test because f 00 (0) = 0. Evaluating the function at the critical points and endpoints, we find that f (−2) = −8, f (0) = 0, and f (2) = 8. Since the first derivative cannot change signs on the subinterval (−2, 0), and f (0) > f (−2), it follows that the first derivative is positive on (−2, 0) (this is an intuitive result, which can be shown rigorously using the mean value theorem). Likewise, the first derivative is positive on (0, 2), so f (0) is not an extreme value. Considering the second derivative, we see that for x < 0 we have f 00 (x) < 0, so f (x) is concave down. For x > 0 we have f 00 (x) > 0, so f (x) is concave up. Since the second derivative exists and changes signs around 0, this is an inflection point, as can be seen above. First we plot the endpoints and the critical point, and use our knowledge of the second derivative to connect the dots. The observation in the above example is a useful one. In order to sketch a function we need to evaluate the function at its endpoints and critical points. In doing so, we can already determine the sign of the first derivative over the subintervals created by the endpoints and the critical points. The mean value theorem tells us that if the difference between the endpoints is positive, there will be some point in the interval with a positive derivative, but since the sign of the derivative cannot change over the entire interval, the first derivative must be positive over the entire interval. We can make similar deductions if the difference of the function values at the endpoints is negative or zero. Example 2 Sketch f (x) = x2 on [−3, 3] using the first and second derivatives. Solution By repeated applications of the power rule, we find that f 0 (x) = 2x, and f 00 (x) = 2. We have a critical point at x = 0, and evaluating the function at the critical points and endpoints we find f (−3) = 9, f (0) = 0, and f (3) = 9. It follows that the function is decreasing on (−3, 0) and increasing on (0, 3), which is cofirmed by looking at the first derivative. It follows f (0) is a local minimum. Also, using the second derivative test we see f 00 (0) > 0, which implies f (0) is a local minimum. In fact, fFor all x, the second derivative f 00 (x) > 0, so the function f (x) is always concave up. Now we plot the endpoints and critical points, connecting the dots with the knowledge that the function is always concave up. Example 3 Find f 0 (x) and f 00 (x) if f (x) = 2x. Use this to analyze the familiar graph of the function. Solution By repeated applications of the power rule, we find that f 0 (x) = 2, and f 00 (x) = 0. For all x, the first derivative f 0 (x) > 0, so the function f (x) is always increasing. Also, for all x, the second derivative is 0. This corresponds to a graph that does not have any concavity, which is the familiar line. x on [−5, 5] using the first and second derivatives. Example 4 Sketch f (x) = x−1 Solution The first thing to notice is that we have a division by 0 at x = 1, which corresponds to a possible vertical asymptote. In this case we should calculate the one-sided limits to look at the behavior of the function on each side of x = 1. We find that lim x→1− x = −∞ and x−1 lim x→1+ x =∞ x−1 Next, we need to calculate the first and second derivatives, using the quotient rule. Let u(x) = x and v(x) = x − 1. We find that u0 (x) = 1 = v 0 (x), as these are both linear functions. We find the derivative as f 0 (x) = v(x)u0 (x) − u(x)v 0 (x) x−1−x −1 = = v(x)2 (x − 1)2 (x − 1)2 In order to calculate the second derivative, we once again need to use the quotient rule. Let u(x) = −1 and v(x) = (x − 1)2 = x2 − 2x + 1. We find that u0 (x) = 0 and v 0 (x) = 2x − 2. Using the quotient rule we find f 00 (x) = v(x)u0 (x) − u(x)v 0 (x) (x − 1)2 · 0 − (−1)(2x − 2) 2(x − 1) 2 = = = 2 2 2 4 v(x) ((x − 1) ) (x − 1) (x − 1)3 2 −1 00 In summary, f 0 (x) = (x−1) 2 and f (x) = (x−1)3 . Now that we’ve completed the arduous task of calculating the above derivatives, we can use them to analyze the behavior of the function. Since both the first and second derivative are fractions in which the numerator is constant, neither can equal zero. Thus, our only point of interest is x = 1, where the first and second derivative are not defined, because the function itself is not defined at x = 0. Since the denominator is nonnegative, the first derivative is negative for all x 6= 1. This indicates that for all x 6= 1 the function f (x) is decreasing. The behavior of f 00 (x) is not so complicated, and it shows us that for x < 1 the function is concave down, as f 00 (x) < 0. For x > 1, f 00 (x) > 0, so the function is concave up. Even though the sign of the second derivative changes around x = 1, since the second derivative is not defined there, we do not call x = 1 an inflection point. Evaluating the function at the end points, we see f (−5) = 5/6 and f (5) = −5/4. Even with this much information, it is difficult to construct an accurate graph of this more complicated function. We could try and do some more in deptch analysis of the first derivative, but we might as well just plot some additional points. We find that f (0) = 0 and f (2) = 1. 6 4 x/(x-1) 2 0 -2 -4 -6 -4 -2 0 x 2 4 Example 5 Sketch f (x) = 2x3 − 7x2 + 5 Solution The first step is to find the first and second derivatives. In this case, we simply apply the power rule to find that f 0 (x) = 6x2 − 14x and f 00 (x) = 12x − 14. The first derivative is defined for all x, so in order to find the critical points let us solve for f 0 (x) = 0 0 = 6x2 − 14x = x(6x − 14) 7 0 Thus we have critical points when x = 0 and x = 14 6 = 3 = 2.3. These points divide f (x) into 3 intervals; within each we need to check the sign of the derivative (one point per interval is sufficient, as the derivative only changes sign at a critical point). By looking at f 0 (x), we can see that for very large or small values of x, the positive quadratic term will dominate the negative linear term, making the first derivative positive, so the function is increasing in the first and last interval. Now we can use a sample point of x = 1, which lies between the two critical points, to find that f 0 (1) = 6−14 = −8 < 0. Thus, in the middle interval the function is decreasing. In order to search for possible inflection points, we set the second derivative equal to 0, and solve 7 for x. We find that 0 = 12x − 14, so x = 14 12 = 6 = 1.16. Now we must compare the sign of the second derivative for x > 1.16 and x < 1.16 to see if it an inflection point or not. We find that for x < 1.16 we have f 00 (x) < 0, so the function is concave down. For x > 1.16 we have f 00 (x) > 0, so the function is concave up. Thus, it is truly an inflection point, and we know the concavity of the function changes from down to up at this point. The final step is to plot the function at the above points of interest: (0, f (0)), (1.16, f (1.16)), and (2.3, f (2.3)), which correspond to (0, 5), (1.16, −1.35), and (2.3, −7.7). Once we have plotted these key points, we simply need to connect the dots, and be cognizant of the concavity and whether or not the function is increasing or decreasing. The final graph is pictured below. 10 2x^3 - 7x^2 + 5 5 0 -5 -10 -2 -1 0 1 x 2 3 4