CALCULUS I MATH 181
Popular in Course
Popular in Mathematics (M)
verified elite notetaker
This 9 page Class Notes was uploaded by Hoyt Beer on Sunday October 25, 2015. The Class Notes belongs to MATH 181 at University of Nevada - Las Vegas taught by Bellomo in Fall. Since its upload, it has received 31 views. For similar materials see /class/228626/math-181-university-of-nevada-las-vegas in Mathematics (M) at University of Nevada - Las Vegas.
Reviews for CALCULUS I
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 10/25/15
Math 1 81 Chapter 4 Review UNIVERSITY OF NEVADA LAS VEGAS Please email with any questions Typos are not a frequent occurrence but can happen 1 Answers will vary absolute max local max absolute mm local mm 2 Critical numbers occur when the derivative is zero or undefined f x 24x6 x 624 1 2 x3 2 1 fx x23 x53WWWx2 x0 2 f xL 2xl leyoyl 3x2 x23 3x23x123 2 3 The function is continuous on the interval given l2 x 2x 2 x2 4 x2 l 2 22 x2 fx 24 x 2 2 4 2x 2 24 x 4 x 4 x 4 x x CP 4 x20 xziz and 2 x220 xzi f 1 lm 1732 f20 fx52 The absolute max is at sqrt2 the absolute min is at 1 Note we throw out the other points because they are outside our interval 4 f is continuous on 02 and differentiable on 02 O l and 2 9 There is a c in 02 where 91 390 5 f 270 f39x3xz125 3x24 xrl15 CllSE02 5 local max in ection point increasing decreasing localmjn 6 f x 76x 3xZ 3xx7 2 f x766x6xil f 7 7 7 o f o777770 0 l 2 increasing forx lt 0 andx gt 2 decreasing for x between 0 and 2 concave up forx gt 1 concave down for x lt 1 critical points at 0 and 2 in ection point at l 1 sec2 x 1 1 a 11m 11m H0 Slnx H0 cos x 1 b m 2 Using L Hopitals Rule 2x 3 2x 3 1 2x 3 ln2x5 limln lim 2x1ln lim 2x5 Hm 2x 5 Hoe 2x1 1 2x5 22x5 22x 3 1 16 Z Z zlim 2x 3 2x43 zlim 2x3 2x251im 16 2x1 W 22x1 Hm 22x1 Ho 2 2x 32x5 2x3 2X1 2X1 lim limeX In eX 8 Him M M gt lx lx 71 39 climxem 1li e 711lim 1 Hoe Hoe x Hoe x 39 We can use L Hopital because when we try to evaluate we get 00 sinx 9 For the function fx 1 cos x a Find the rst and second derivatives fx71cosxcosxsin2x7 cosx1 7 1 T 1cosx2 T 1cosx2 T 1cosx f x 711 cos x 27sin x L 1 cos x b The first derivative is unde ned when cosine is negative 1 or i2k17r for k 012 The rst derivative is never zero and is always positive when de ned The second derivative is undefined at the same place as above i2k 17r The second derivative is zero when sine is zero or i2k7r f u70 1170 u70u fru u u u 3n 2n 17r 07 In 21 31 c Critical points i2k17r In ection points i2k7r d fquotu0 1170 ueou7 f u u u u 31t 21 11 01 In 21 31E 10 If 1000 sq cm of material is available to make a box with a square base and an open top find the largest possible volume of the box Box dimensions base x height 11 Volume ofbox xzh 1000 7 x2 4x Surface area of box bottom 4 sides x2 4x11 21000 So 11 Volume x2 1002 sz 0000 x Differentiating we nd 10007 3x2 2 0 x 1826 The second derivative is always negative so this is indeed a max 2 h 100071826 913 41826 To maximize the volume the box should have a base of 1826 cm and a height of9 13cm 23x2 x 11 P x R x C x x10 6503x005x2 7x 650 210 420 46x Differentiating we nd P x 7 m50 3 x6391 This is a max because the second derivative is always negative concave down 12 x4 4x3 7x2 20x2339 4x3 12x2 14x 20 x2xl amp amp2 2190625 f190625035 fx1 f2 3 x3x2 M190625 M1892015 f18920150007 fx2 f190625 So the root is approximately 1892015 13 gx 20x7412 7x 4xc 5x4 493 gx2 4xc g2 1 implies that g2 52 423 522 42c 21 134C13 C 133 So gx 5x4 493 x2 4x 133 Math 1 81 Chapter 2 Review UNIVERSITY OF NEVADA LAS VEGAS This is not a binding contract only a guideline for study Questions are given in entirety or outlined If applicable the section number and block of exercises similar to it are given in parenthesis Find the limit lim xgt74 Find the limit lim xgt0 x2 Find the limit lim 2 xgt0 x Evaluate lim xgt3 Find all asymptotes horizontal and vertical for f x Describe what is meant by a tangent line and sketch an example of one x2 16 x4 Vx25 5 x 2 x 2x 2x and HO x x l xl x2 9 lx3l by evaluating the limit from the right and left Prove lin314x 2 10 using the epsilondelta definition of a limit Sketch a function that has at least one removable discontinuity and at least one infinite discontinuity and at least one jump discontinuity Be sure to label the discontinuities Prove that f x x tanx has a root between x 75 and 78 3x22x 5 2x2 4 F1nd the slope of the tangent to the curve y 1 at any pomt us1ng the hnut def1mtlon of x derivative Math 181 Chapter 2 Review Solutions UNIVERSITY OF NEVADA LAS VEGAS Please email with any questions Typos are not a frequent occurrence but can happen 1 Describe what is meant by a tangent line and sketch an example of one The tangent line touches the curve only one point and has the same direction as the curve x2 16 2 Find the limit lin41 x4 hm x4x 4 hm x 4 xgt74 x 4 xgt74 1 3 Find the limit ling x255 xgt x hmxx25 5x255hm x25 25 hm l 1 H0 x x255 H0xx255 H0lx255 10 2 2 4 Find the limit lime zx and lime zx xgt0 x x l xeoi x XS 1 x Z llmz z llm oo xao x 05 1 xgt0 xx 1 x2 2x x Z lim 2 2 11m HO x x 1 HO xx l 2 r 3939 by evaluating the limit from the right and left x 5 Evaluate lim H m x73x3 m 93 HT 6 HT 73 x73x37 4x73 lim Hr lim Hr x So lim 9 DNE Hz Ix 3 6 Prove lin 4x r 2 10 using the epsilondelta de nition ofa limit H Let 5 gt 0 begiven and assume that x73 lt 6 for some 5 gt 0 Remember it is our job to nd that delta fx7L4x727104x7124x73 Wewantto force fxrL ltsso 4xr3 lt46 lt5 so we require that 6 lt 7 ketcha 39 439 39 andat 39 439 quot quot 39 in nite 8 8 Prove that fx x itanx has a root between x 75 and 78 f75 75rtan75 4794 gt 0 f78 787 mus 710707 lt 0 Since the c 39 39 39 4 439 value theorem it has a root between 75 and 78 39LI 7578by 39 2 9 Find all asymptotes horizontal and vertical for f x x 3 2 5 3 2 3 llm 11m 2x2 So it has a horizontal asymptote at f 32 Hoe x Hoe x Z Z w So the function is unde ned at ixE 2x 2 2xJ x J Since 3x2 2x 5 0 when x ixE these are vertical asymptotes 10 Find the slope of the tangent to the curve y 1 at any p01nt us1ng the 11m1t de nltion of x derivative 1 1 11 1 i x xh r xh x 1 xh x21 xxh xxh1 h hao h hao h hao h hgt0 hxxh 1 1 lim HO xx h x2 l The slope of the tangent at any p01nt x 1s given by 2 x
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'