Chapter 4 22 - 234 CHAPTER 4 APPLICATIONS OF...

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: 234 CHAPTER 4 APPLICATIONS OF DlFFERENTIATlUN 2m3/2(#1/at) —- (2 — 1n m)(3a:1/2) _ —2:1:1/2 + 3w1/2(ln:c * 2) (C) f"(IE) = (2533/2)2 41:3 _a:1/2(—2+3lnm-6) 3111:5—8 4353 — 4955/2 f”(m) = 0 41> 1111: 2% (i) m : 88/3. f"(:L') >0 (I) w > 68/3. so f is concave upward on (68/3. 00) and concave downward on (0. (38/3). There is an inflection point at (es/3. §e_4/3) m (14.39. 0.70). 20. (a) y : f(rn) : xlnrc. (Note that f is only defined fora: >0.) f'(1:) : 33(1/31) +lnm :1+ln$. f'($) > 0 <:> lnm+ 1 > 0 4:} lnm > —1 <:> yr > 6’1. Therefore f is increasing on (1/6. 00) and decreasing on (0. 1/6). (b) f changes from decreasing to increasing at a: : 1/e. so f(1/e) : #1/6 is a local minimum value. (C) f"(ac) : 1/3: > 0 for m > 0. So f is concave upward on its entire domain. and has no inflection point, 21. f(a:)=a:5r5$+3 => f’(m):5m475:5(m2+1)($+1)(:c~1). First Derivative Test: f'(a:) < 0 :> *1 < a: < 1 and f'(:c) > 0 :> a: > 1 orm < —1. Since f' changes from positive to negative at :c : -1. f (—1) : 7 is a local maximum value; and since f’ changes from negative to positive at :c : 1, f(1) : —1 is a local minimum value. Second Derivative Test: f”($) : 20x3. f’(a:) : 0 <=> w : ::1. f"(*1) : 720 < 0 :> f(#1) : 7 is a local maximum value. f"(1) : 20 > 0 => f(1) : —1 is a local minimum value. Preference: For this function. the two tests are equally easy. ac , (m2+4)-1#$(2$) 4~$2 (2+$)(2—:E) 22' f(m) A $32 + 4 f (m) g (as? + 4)2 _ (132 + 4)2 ($2 + 4)2 ' vative Test: f'(m) > 0 :> —2 < ac < 2 and f'(:c) < O :> at > 2 ora: < —2. Since f’ changes First Deri from positive to negative at w : 2. f (2) : i is a local maximum value; and since f’ changes from negative to positive at a: : —2, f(—2) : 4% is a local minimum value. Second Derivative Test: (a:2 + 4)2(~2m) 7 (4 — $2) - 2(332 + 4)(2:c) // m ’ f ( ) W + 4W #2m(m2 + 4) [(m2 + 4) + 2(4 4 232)] # 4237(12 4 3:2) ' (m2 + 4)4 (562 + 4):; f’(a:) : 0 4:) a; : :l:2. f”(#2) : fi > 0 :> f(—2) : ~i is alocal minimum value. f"(2) : 7%5 < 0 => f(2) : 711 is alocal maximum value. Preference: Since calculating the second derivative is fairly difficult. the First Derivative Test is easier to use for this function. ...
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