Math 3, Section 5 (Troyka) Fall 2016 Solution to Group Work October 14 (Class 15)

Math 3, section 5 (Troyka) Fall 2016 Solution to Group Work October 14 (class 15)

Using the techniques we have learned, ﬁnd all of the important features of the graph of the given function. Then draw a careful, detailed sketch of the graph. For each one, we will ﬁnd the following data about the function:

(a) intercepts

(b) asymptotes

(d) local minima and maxima

(e) intervals where the function is concave up/down

(f) inﬂection points

Because I am typing this, it’s difficult for me to include diagrams and pictures. In your solution, you should include the diagram of the number line with all the critical points, showing where the derivative is positive versus negative; you should do a similar thing with the second derivative and the potential points of inflection. For the same reason, I’m also not including sketches. You can use a graphing calculator or an internet tool to check your sketch. t2 1. g(t) = . t + 3 Summary. The y-intercept is y = 0, and the t-intercept is t = 0. There is a vertical asymptote at t = −3, where the curve goes down to −∞ on the left and goes up to +∞ on the right. There are no horizontal asymptotes, but g goes to −∞ on the left and +∞ on the right. The function starts out increasing for t < −6, until it reaches a local maximum with coordinates (−6, −12); then it decreases for −6 < t < −3, until the vertical asymptote at t = −3; then it decreases again for −3 < t < 0, until it reaches a local minimum with coordinates (0, 0); then it increases for t > 0. The function is concave down before the vertical asymptote, concave up after; it has no inflection points. (a) The y-intercept is g(0) = 0/(0 + 3) = 0. The t-intercepts are the points where g(t) = 0: t2 = 0 ⇔ t2 = 0 ⇔ t = 0. t + 3 So t = 0 is the only t-intercept.

1 (b) There is a vertical asymptote at points where the limit is infinity from one or both sides. The only place where this can happen is t = −3, because g is a rational function and hence it is continuous everywhere except where the bottom of the fraction is 0. We check whether t = −3 is a vertical asymptote by taking the limit from each side: t2 t2 ∼ 32 lim = −∞ because → ; t→−3− t + 3 t + 3 small negative number t2 t2 ∼ 32 lim = +∞ because → t→−3+ t + 3 t + 3 small positive number Thus t = −3 is a vertical asymptote, and the curve goes down to −∞ on the left of the asymptote and goes up to +∞ on the right of the asymptote. To check for horizontal asymptotes, we look at the limits to ±∞: t2 t2 1/t t lim = lim · = lim = −∞; t→−∞ t→−∞ t→−∞ 3 t + 3 t + 3 1/t 1 + t t2 t2 1/t t lim = lim · = lim = ∞. t→∞ t→∞ t→∞ 3 t + 3 t + 3 1/t 1 + t (In both cases, we reach an expression where the top is t and the bottom is going to 1.) Thus there are no horizontal asymptotes, but the function goes to −∞ on the left and +∞ on the right. (c) To find where the function is increasing/decreasing, we find the critical points. We find g0 using the quotient rule: (t2)0(t + 3) − t2(t + 3)0 2t(t + 3) − t2(1) (2t2 + 6t) − t2 t2 + 6t g0(t) = = = = . (t + 3)2 (t + 3)2 (t + 3)2 (t + 3)2

t2 + 6t So g0(t) = . The critical points are where g0 is zero or undeﬁned. It is undeﬁned (t + 3)2 where the bottom of the fraction is 0, which is at t = −3. It is zero where the top is 0 (and the bottom is not −3):

t2 + 6t = 0 ⇔ t(t + 6) = 0 ⇔ t = −6, 0.

Therefore the critical points are −6, −3, and 0. To tell whether g is increasing or decreasing, we check whether g0 is positive or negative between the critical points. (−100)2 + 6(−100) (−4)2 + 6(−4) g0(−100) = > 0; g0(−4) = = −8 < 0; (−100 + 3)2 (−4 + 3)2 (−1)2 + 6(−1) 12 + 6(1) g0(−1) = = −5/4 < 0; g0(1) = = 7/16 > 0. (−1 + 3)2 (1 + 3)2

2 So g0 is positive on the intervals (−∞, −6) and (0, ∞), and g0 is negative on (−6, −3) and (−3, 0). Therefore, g is increasing on (−∞, −6) and (0, ∞), and g is decreasing on (−6, −3) and (−3, 0). (d) To determine the local maxima and minima of g, we must classify each critical point as a local minimum, local maximum, or neither. The critical points are −6, −3, and 0; but −3 is not in the domain of g, so it cannot be a local extremum. For the two remaining critical points, we can use the first derivative test: g0 is positive on (−∞, −6) and negative on (−6, −3), so the critical point at −6 is a local maximum; similarly, g0 is negative on (−3, 0) and positive on (0, ∞), so the critical point at 0 is a local minimum. The y-value of the local maximum at −6 is g(−6) = −12, and the y-value of the local minimum at 0 is g(0) = 0. Therefore, g has a local maximum with coordinates (−6, −12), and g has a local minimum with coordinates (0, 0). (e) To find where the function is concave up/down, we take the second derivative. Differentiating g0, we again use the quotient rule:

(t2 + 6t)0(t + 3)2 − (t2 + 6t) ((t + 3)2)0 g00(t) = ((t + 3)2)2 (2t + 6)(t + 3)2 − (t2 + 6t)(2)(t + 3) = (t + 3)4 (2t + 6)(t + 3) − 2(t2 + 6t) 18 = = . (t + 3)3 (t + 3)3 18 So g00(t) = . The points where g may change concavity are the points where g00 is (t + 3)3 undefined or 0. It is undefined where the bottom of the fraction is 0, which is at t = −3. It is never zero, because the top of the fraction is 18. Therefore the concavity can change direction only at −3. To tell whether g is concave up or down, we check whether g00 is positive or negative between the points where it may change concavity: 18 18 2 g00(−100) = < 0; g00(0) = = > 0. (−100 + 3)3 (0 + 3)3 3 So g00 is negative before −3 and positive after −3. Therefore, g is concave down on (−∞, −3) and concave up on (−3, ∞). (f) The inflection points, by definition, are the points where g changes between concave up and concave down. We saw that this can only happen at −3, but this is not in the domain of g. Therefore g has no inflection points. 2. f(x) = (x2 + 2x + 2)e−x. Summary. The y-intercept is y = 2, and there is no x-intercept. There are no vertical asymptotes. There is a horizontal asymptote at y = 0 on the right of the curve, and f

3 goes to +∞ on the left. The function is decreasing everywhere, although it has a horizontal tangent line at a single point at x = 0; there are no local extrema. The function starts out concave up for x < 0, until it reaches an inﬂection point with coordinates (0, 2); then it is concave down for 0 < x < 2, until it reaches an inﬂection point with coordinates (2, 10/e2); then it is concave up for x > 2. (a) The y-intercept is f(0) = (02 + 2(0) + 2)e0 = 2. The x-intercepts are the points where f(x) = 0:

(x2 + 2x + 2)e−x = 0 ⇔ x2 + 2x + 2 = 0 or e−x = 0.

But x2 + 2x + 2 is always strictly positive, and so is e−x. Therefore f is always positive and has no x-intercept. (b) There is no vertical asymptote: f is continuous everywhere, since it is a product of a polynomial and an exponential function, both continuous functions. To check for horizontal asymptotes, we look at the limits to ±∞:

lim (x2 + 2x + 2)e−x = ∞ x→−∞ because lim (x2 + 2x + 2) = ∞ and lim e−x = ∞. Next, x→−∞ x→−∞

x2 + 2x + 2 lim (x2 + 2x + 2)e−x = lim = 0, x→∞ x→∞ ex because ex grows much larger than any polynomial as x → ∞. Therefore f has a horizontal asymptote at y = 0 on the right, and f goes to ∞ on the left. (c) To find where the function is increasing/decreasing, we find the critical points. We find f 0 using the product rule and chain rule:

f 0(x) = (x2 + 2x + 2)0e−x + (x2 + 2x + 2) e−x0 = (2x + 2)e−x + (x2 + 2x + 2)(−e−x) = (2x + 2 − x2 − 2x − 2)e−x = −x2e−x.

So f 0(x) = −x2e−x. The critical points are where f 0 is zero or undefined. It is never undefined, because it is the product of a polynomial and an exponential function, both defined everywhere. So the critical points are just where f 0(x) = 0:

−x2e−x = 0 ⇔ −x2 = 0 or e−x = 0.

But e−x is always strictly positive, never zero. So f 0(x) = 0 when −x2 = 0, and the solution is just x = 0. Therefore, the only critical point is x = 0. To tell whether f is increasing or decreasing, we check whether f 0 is positive or negative between the critical points.

f 0(−1) = −(−1)2e1 = −e < 0; f 0(1) = −(12)e−1 = −1/e < 0.

4 So f 0 is negative on both the intervals (−∞, 0) and (0, ∞), and f 0 is never positive. Therefore, f is decreasing on (−∞, 0) and on (0, ∞), and f is never increasing. (In fact, we could say f is decreasing at 0 too, since it’s decreasing on both sides of 0; in this case f is decreasing everywhere.) (d) To determine the local maxima and minima of f, we must classify the one critical point as a local minimum, local maximum, or neither. The only critical point is 0, and we now apply the first derivative test: f 0 is negative on (−∞, 0) and also negative on (0, ∞), so 0 is neither a local maximum nor a local minimum (since f 0 does not change between positive and negative at 0). Therefore f has no local extrema. (e) To find where the function is concave up/down, we take the second derivative. Differentiating f 0, we again use the product rule and chain rule:

f 00(x) = (−x2)0e−x + (−x2) e−x0 = −2x e−x − x2(−e−x) = (x2 − 2x)e−x.

So f 00(x) = (x2 − 2x)e−x. The points where f may change concavity are the points where f 00 is undefined or 0. It is never undefined, because it is the product of a polynomial and an exponential function, both defined everywhere. So the concavity can only change direction when f 00(x) = 0: (x2 − 2x)e−x = 0 ⇔ x2 − 2x = 0 or e−x = 0. But e−x is always strictly positive, never zero. So f 0(x) = 0 when x2 − 2x = 0. This factors as x(x − 2) = 0, so the solution is x = 0 or 2. Therefore the concavity can change direction only at 0 and 2. To tell whether f is concave up or down, we check whether f 00 is positive or negative between the points where it may change concavity:

f 00(−1) = ((−1)2 − 2(−1))e1 = 3e > 0; f 00(1) = (12 − 2(1))e−1 = −1/e < 0; f 00(3) = (32 − 2(3))e−3 = 3/e3 > 0.

So f 00 is positive on the intervals (−∞, 0) and (2, ∞), and f 00 is negative on the interval (0, 2). Therefore, f is concave up on (−∞, 0) and (2, ∞), and f is concave down on (0, 2). (f) The inflection points, by definition, are the points where f changes between concave up and concave down. We saw that this can only happen at 0 and 2. Since f is concave up before 0 and concave down on (0, 2), it changes concavity at 0; similarly, since f is concave down on (0, 2) and concave up after 2, it changes concavity at 2. Therefore, the inflection points are 0 and 2. The y-value of the inflection point at 0 is f(0) = 2, and the y-value of the inflection point at 2 is f(2) = 10/e2. Therefore, the inflection points of f have coordinates (0, 2), and (2, 10/e2). (The latter point is at approximately (2, 1.35).) 12x2 + 36 3. h(x) = . (x + 3)2

5 Summary. The y-intercept is y = 4, and there is no x-intercept. There is a vertical asymptote at x = −3, where the curve goes up to +∞ on the left and the right. There is a horizontal asymptote at y = 12, on the left of the curve and the right of the curve. The function starts out increasing for x < −3, until the vertical asymptote at x = −3; then it decreases for −3 < x < 1, until it reaches a local minimum with coordinates (1, 3); then it increases again for x > 1. The function starts out concave up for x < −3, until the vertical asymptote at x = −3; then it is concave up again for −3 < x < 3, until it reaches an inﬂection point with coordinates (3, 4); then it is concave down for x > 3. (a) The y-intercept is h(0) = 36/32 = 4. The x-intercepts are the points where h(x) = 0: 12x2 + 36 = 0 ⇔ 12x2 + 36 = 0, (x + 3)2 but 12x2 + 36 has no real roots. Therefore h has no x-intercept. (b) There is a vertical asymptote at points where the limit is inﬁnity from one or both sides. The only place where this can happen is x = −3, because h is a rational function and hence it is continuous everywhere except where the bottom of the fraction is 0. We check whether x = −3 is a vertical asymptote by taking the limit from each side: 12x2 + 36 12x2 + 36 ∼ 12 · 32 + 36 lim = +∞ because → ; x→−3− (x + 3)2 (x + 3)2 small positive number 12x2 + 36 12x2 + 36 ∼ 12 · 32 + 36 lim = +∞ because → . x→−3+ (x + 3)2 (x + 3)2 small positive number Thus x = −3 is a vertical asymptote, and the curve goes up to +∞ on the left and right of the asymptote. To check for horizontal asymptotes, we look at the limits to ±∞:

2 2 2 36 12x + 36 12x + 36 1/x 12 + 2 12 lim = lim · = lim x = = 12; x→−∞ 2 x→−∞ 2 2 x→−∞ 6 9 (x + 3) x + 6x + 9 1/x 1 + x + x2 1 2 2 2 36 12x + 36 12x + 36 1/x 12 + 2 12 lim = lim · = lim x = = 12. x→∞ 2 x→∞ 2 2 x→∞ 6 9 (x + 3) x + 6x + 9 1/x 1 + x + x2 1 Thus there is a horizontal asymptote at y = 12, on the left and on the right. (c) To find where the function is increasing/decreasing, we find the critical points. We find h0 using the quotient rule: (12x2 + 36)0(x + 3)2 − (12x2 + 36) ((x + 3)2)0 h0(x) = ((x + 3)2)2 24x(x + 3)2 − (12x2 + 36)(2)(x + 3) = (x + 3)4 24x(x + 3) − 2(12x2 + 36) 72x − 72 = = . (x + 3)3 (x + 3)3

6 72x − 72 So h0(x) = . The critical points are where h0 is zero or undefined. It is undefined (x + 3)3 where the bottom of the fraction is 0, which is at x = −3. It is zero where the top is 0 (and the bottom is not −3): 72x − 72 = 0 ⇔ x − 1 = 0 ⇔ x = 1. Therefore the critical points are −3 and 1. To tell whether h is increasing or decreasing, we check whether h0 is positive or negative between the critical points. 72(−100) − 72 −72 h0(−100) = > 0; h0(0) = = −2 < 0; (−100 + 3)3 33 72(100) − 72 h0(100) = > 0. (100 + 3)3 So h0 is positive on the intervals (−∞, −3) and (1, ∞), and h0 is negative on (−3, 1). There- fore, h is increasing on (−∞, −3) and (1, ∞), and h is decreasing on (−3, 1). (Notice that we needed to count −3 as a critical point in order to catch the change from increasing to decreasing at −3.) (d) To determine the local maxima and minima of h, we must classify each critical point as a local minimum, local maximum, or neither. The critical points are −3 and 1; but −3 is not in the domain of h, so it cannot be a local extremum. For the one remaining critical point, we can use the first derivative test: h0 is negative on (−3, 1) and positive on (1, ∞), so the critical point at 1 is a local minimum. The y-value of the local minimum at 1 is h(1) = 3. Therefore, h has a local minimum with coordinates (1, 3). (e) To find where the function is concave up/down, we take the second derivative. Differentiating h0, we again use the quotient rule: (72x − 72)0(x + 3)3 − (72x − 72) ((x + 3)3)0 h00(x) = ((x + 3)3)2 72(x + 3)3 − (72x − 72)(3)(x + 3)2 = (x + 3)6 72(x + 3) − 3(72x − 72) −144(x − 3) = = . (x + 3)4 (x + 3)4 −144(x − 3) So h00(x) = . The points where h may change concavity are the points where h00 (x + 3)4 is undefined or 0. It is undefined where the bottom of the fraction is 0, which is at x = −3. Next we find the points where h00(x) = 0: −144(x − 3) = 0 ⇔ −144(x − 3) = 0 ⇔ x = 3. (x + 3)4

7 Therefore the concavity can change direction at −3 and 3. To tell whether h is concave up or down, we check whether h00 is positive or negative between the points where it may change concavity:

−144(−100 − 3) −144(−3) h00(−100) = > 0; h00(0) = > 0; (−100 + 3)4 34 −144(100 − 3) h00(100) = < 0. (100 + 3)4

So h00 is positive on the intervals (−∞, −3) and (−3, 3), and h00 is negative on the interval (3, ∞). Therefore, h is concave up on (−∞, −3) and (−3, 3), and h is concave down on (3, ∞). (f) The inflection points, by definition, are the points where h changes between concave up and concave down. We saw that this can only happen at −3 and 3. But −3 is not in the domain of f, so it cannot be an inflection point. As for 3, since h is concave up before 3 and concave down after 3, it changes concavity at 3. Therefore, 3 is the only inflection point. The y-value of the inflection point at 3 is h(3) = 4. Therefore, the inflection point of h has coordinates (3, 4). 4. p(x) = x(1 + x2)−1/2. x Alternatively, we can write this function in the form p(x) = √ . 1 + x2 Summary. The y-intercept is y = 0, and the x-intercept is x = 0. There is no vertical asymptote. There are two horizontal asymptotes: one at y = −1 on the left of the curve, and another at y = 1 on the right of the curve. The function is increasing everywhere, and it has no local extrema. The function is concave up for x < 0, has an inflection point with coordinates (0, 0), and is concave down for x > 0. (a) The y-intercept is p(0) = 0. The x-intercepts are the points where p(x) = 0:

x √ = 0 ⇔ x = 0. 1 + x2

Therefore p has x-intercept x = 0.

(b) There is a vertical asymptote√ at points where the limit is inﬁnity from one or both sides. Observe that x and 1 + x2 are continuous everywhere (the expression under the x square root, 1 + x2, is always positive), and so the fraction p(x) = √ is continuous 1 + x2 everywhere except where the bottom of the fraction is 0. But the bottom of the fraction is never zero! Thus p has no vertical asymptotes. To check for horizontal asymptotes, we look at the limits to ±∞. We will use the fact

8 √ that x2 = |x|, and the fact that x/|x| = −1 for x < 0 and x/|x| = 1 for x > 0:

x x x 1/|x| |x| −1 −1 lim √ = lim √ · √ = lim q = lim q = = −1; x→−∞ 1 + x2 x→−∞ 1 + x2 1/ x2 x→−∞ 1 x2 x→−∞ 1 1 x2 + x2 x2 + 1 x x x 1/|x| |x| 1 1 lim √ = lim √ · √ = lim q = lim q = = 1. x→∞ 1 + x2 x→∞ 1 + x2 1/ x2 x→∞ 1 x2 x→∞ 1 1 x2 + x2 x2 + 1

Thus there is a horizontal asymptote at y = −1 on the left of the curve, and there is a different horizontal asymptote at y = 1 on the right of the curve. (c) To find where the function is increasing/decreasing, we find the critical points. We find p0 using the product rule and the chain rule, then simplifying a lot:

0 p0(x) = (x)0(1 + x2)−1/2 + x (1 + x2)1/2 (product rule) 2 −1/2 1 2 −3/2 2 0 = 1(1 + x ) + x · (− 2 )(1 + x ) · (1 + x ) (chain rule) 2 −1/2 1 2 −3/2 = (1 + x ) − x · 2 (1 + x ) · 2x = (1 + x2)−1/2 − x2(1 + x2)−3/2 1 x2 = − (and now we simplify . . . ) (1 + x2)1/2 (1 + x2)3/2 1 1 + x2 x2 = · − (1 + x2)1/2 1 + x2 (1 + x2)3/2 1 + x2 x2 = − (1 + x2)3/2 (1 + x2)3/2 1 = . (1 + x2)3/2

1 So p0(x) = . The critical points are where p0 is zero or undefined. Because there is (1 + x2)3/2 a square root involved, p0 is undefined when the expression in the square root is negative — but 1 + x2 is always positive, so this does not occur. The other way that p0 can be undefined is for the bottom of the fraction to be zero — but 1 + x2 is never zero, so this does not occur. Therefore, p0 is never undefined. Also, p0 is never zero, since the top of the fraction is 1. Therefore, there are no critical points! In fact, p0(x) is positive for all x, since (1 + x2)3/2 is taking the positive square root. Therefore, p is increasing everywhere. (d) There are no critical points, so there are no local extrema. (e) To find where the function is concave up/down, we take the second derivative.

9 Diﬀerentiating p0, we use the chain rule:

00 2 −3/20 3 2 −5/2 20 p (x) = (1 + x ) = (− 2 )(1 + x ) · 1 + x 3x = (− 3 )(1 + x2)−5/2 · 2x = − . 2 (1 + x2)5/2 3x So p00(x) = − . The points where p may change concavity are the points where (1 + x2)5/2 p00 is undefined or 0. Again, 1 + x2 is always strictly positive, so the square root is always defined, and dividing by (1 + x2)5/2 is always defined. So the concavity can only change direction when p00(x) = 0, which happens when x = 0. Therefore the concavity can change direction only at 0. To tell whether p is concave up or down, we check whether p00 is positive or negative between the points where it may change concavity:

3(−100) 3(100) p00(−100) = − > 0; p00(100) = − < 0. (1 + (−100)2)5/2 (1 + 1002)5/2

So p00 is positive before 0 and negative after 0. Therefore, p is concave up on (−∞, 0) and concave down on (0, ∞). (f) The inflection points, by definition, are the points where p changes between concave up and concave down. We saw that this can only happen at 0. Since p is concave up before 0 and concave down after 0, it changes concavity at 0. Therefore, 0 is the only inflection point. The y-value of the inflection point at 0 is p(0) = 0. Therefore, the inflection point of p has coordinates (0, 0).