M408C: Recap and Curve SketchingOctober 9, 2008It is worthwhile to review Sections 4.3 (FDT, SDT, etc) and 4.4 (horizontal asymptotes) as wellas moving on to Section 4.5 (curve sketching). Section 4.5 basically requires utilizing all the skillsyou have learned so far to sketch accurate graphs of functions. It requires mastery in just abouteverything that we have been doing so far in class.1.2.3.14. (4.3.38) Find the intervals of increase or decrease, the local max and min, the intervals ofconcavity, and the inflection points for G(x) = x − 4√x.Solution: Observe that the domain of G is x ∈ [0, ∞). G0(x) = 1−4·12x−1/2= 1−2/√x =√x−2√x.Now G0(x) > 0 if and only if√x−2 > 0, i.e. for x > 4. So G is increasing on (4, ∞) and decreasingon (0, 4). We know x = 4 is a critical point of G, and we also know that G0(x) goes from negativeto positive across x = 4. Thus x = 4 is a local minimum with value G(4) = −4.Now G00(x) = 1 − 2 ·−12x−3/2= 1 + x−3/2=√x3+1√x3, and G00(x) > 0 for all x > 0. Hence G isalways concave up and there are no inflection points. Fro this information, you should also be ableto graph G.5. (4.4.36) Find the horizontal and vertical asymptotes for y =1+x4x2−x4.Solution: We write y =1+x4x2(1−x2)=1+x4x2(1+x)(1−x). Thus y has vertical asymptotes at x = 0, 1, −1.We can also see readily thatlimx→−1−y = −∞ limx→0−y = ∞ limx→1−y = ∞limx→−1+y = ∞ limx→0+y = ∞ limx→1+y = −∞To find horizontal asymptotes, considerlimx→∞1 + x4x2− x4·1x41x4= limx→∞1x4+ 11x2− 1= −1 and limx→−∞1 + x4x2− x4·1x41x4= limx→−∞1x4+ 11x2− 1= −1Thus y = −1 is a horizontal asymptote.6. (4.5.10) Sketch the graph of the following function using the guidelines in Section 4.5: y =x2−4x2−2x.Solution: Note that y(x) = y =x2−4x2−2x=(x+2)(x−2)x(x−2).A, Domain: Thus the domain of the function is all x 6= 0, 2, i.e. (−∞, 0) ∪ (0, 2) ∪(2, ∞).B, Intercepts: For x 6= 2, y =x+2x, so I will henceforth assume x 6= 2. Clearly y(0) does notexist, so y has no y-intercept. On the other hand y(x) = 0 if and only if x = −2, so y has anx-intercept of x = −2.C, Symmetry: y(−x) =(−x)2−4(−x)2−2(−x)which is neither y(x) nor −y(x). Thus y has no symmetry.D, Asymptotes: There is a vertical asymptote at x = 0. Remember that I’m assuming x 6= 2,so I can write y =x+2x. If you don’t believe me that there isn’t a vertical asymptote at x = 2,computing the left and right limits there. A quick argument shows that limx→0−x+2x= −∞ andlimx→0+x+2x= ∞. For horizontal asymptotes, we notice thatlimx→∞x + 2x= limx→∞x + 2x·1x1x= limx→∞1 +2x1= 1 = limx→−∞x + 2x.Thus y = 1 is a horizontal asymptote.E, Intervals of Increase or Decrease: y0(x) =x−(x+2)x2= −2/x2= −2x−2. So y0(x) < 0 for allx 6= 0, so y is always decreasing.F, Local Maxima and Minima: y0(x) 6= 0 for any x, so there are no critical points, and henceno local max or min.G, Concavity and Inflection Points: y00(x) = −2(−2x−3) = 4/x3. So y00(x) > 0 for all x > 0and y00(x) < 0 for all x < 0. Thus y is concave up for x > 0 and concave down for x < 0. Theinflection point is just x = 0, for which y is not defined.H. Sketch the Curve: You should be able to do this with the information given