Chapter 4 7 - 57 59 60 61 62.f(t 3/7503 t[0,8 f(t 8161/3...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 57. 59. 60. 61. 62. .f(t) : 3/7503 , t). [0,8]. f(t) : 8161/3 — W3 :> f’(t) : gr?” — §t1/3 : §t*2/3(2 it) : SECTION 4.1 MAXIMUM AND MINIMUM VALUES C 269 f’(t)=t-l(4_t2)‘1/2(—2t)+(4_t2)1/2-1= ‘tz +¢4—t2=flfl:4"2t2- 2 x/4—t2 x/4—t2 x/4—t2 f’(t) = 0 => 4 — 2t2 2 0 : t2 = 2 2 25: ::\/§. butt 2 —\/§ is not in the given interval. [—1.2]. f’(t) does not exist if4 — t2 =0 => t: :l:2. but —2 is not in the given interval. f(~1) : —\/§. f(\/§) = 2, and f(2) = 0. So f(\/§) : 2 is the absolute maximum value and f(—1) : —\/§ is the absolute minimum value. f’(t) : 0 :> t : 2. f’(t) does not exist ift : 0. f(0) : 0, f(2) = 6 W m 7.56. and f(8) : 0. So f(2) : 6 W is the absolute maximum value and f(0) : f(8) : 0 is the absolute minimum value. 'n f($):sinm+cosg;~ [0%] f’(m)=c0sm~sinm:0 {I} sinxzcosa: => :;8:=1 => tanx :1 : a: : g. f(0) : 1.1%) : Vim 1.41.f(§) : fizfl % 1.37. Sofig) : fiis the absolute maximum value and f(0) : 1 is the absolute minimum value. f(1:) : :c * 2cosm. [—n.7r]. f'(ac) : 1 + 231nm = 0 42> sinm = — f(~7r) : 2 — 7r m —1.14.f(—%") : f~ 5?" a: 4.886. f(~%) : 7% NIH _,_ _£ <=> 93* . s- f m —2.26. 3 f(7r) : 71' —l— 2 m 5.14. So f(7r) = 7r + 2 is the absolute maximum value and f(i%) : 7% e \/§ is the absolute minimum value. me) : we. [0.21. f’tr) : w<—e*1>+ 2 Wu — z) : 0 e m = 1. f(0) : 0. f(1) : 6—1 : 1/6 x 0.37. f(2) : 2/e2 % 0.27. So f(1) : 1/8 is the absolute maximum value and f(0) : 0 is the absolute minimum value. l , 1 —1 1—1 16(35):? l1>3l- f($)=w= $2n17_0 $1 lnx—O \—> lnm—l 4:) 11:6. f(1) : 0/1 : 0.f(e):1/e m 0.368. f(3) : (ln 3)/3 m 0.366. So f(e) : 1/6 is the absolute maximum value and f(1) : 0 is the absolute minimum value. E _ a: — 3 :c w in the domain off. f(1) : 1. f(3) : 3 — 3ln3 % —0.296, f(4) : 4 — 3ln4 % 70.159. So f(1) : 1 is the fix) 2 x — 31mm. [1.4]. f’(:c) * 1 ~ ~ 0 7 m ~ 3. f’ does not exist for :c : 0. but 0 is not absolute maximum value and f(3) : 3 — 3111 3 z —0.296 is the absolute minimum value. “spake”? [0.1]. f'(m)~eex( 1) e-‘W 2): 2 1 -2‘ew20 4:) 4:2 4:) 62$ 61 82$ ac : ln2 % 0.69. f(0) : 0. film) : e—‘n2 — e“21n2 : (61“2)‘1 1 (emf2 : 2‘1 — 2*2 : l 2 f(1) : 6—1 — e72 x 0.233. So f(ln 2) : i is the absolute maximum value and f(0) : 0 is the absolute minimum value. ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern