File0024 - 104 25 26 27 28 29 30 31 Chapter 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: 104 25. 26. 27. 28. 29. 30. 31. Chapter 3 Differentiation ~ 1 " — iLi“’—(7“L _ _A___ __ —1 g [7(X) 211m xLfiL. zlgnxLl—L— " zlglxk x x)(z—i)(x—1) 7 znm y x(z x)(zzi+1))((x71) _ ZleX(Z—1)(X— 1) a —1 — (x7117 _ _ (1+\/—)( (1+\/;) _ — fi+fi:1 A g’(X)—zlignx——%‘—’_1im(——ZZT—)—zlfllx% m— .121me :szXW: 7 Note that as x increases, the slope of the tangent line to the curve is first negative, then zero (when x : 0), then positive :> the slope is always increasing which matches (b). Note that the slope of the tangent line is never negative. For x negative, fé(x) is positive but decreasing as x increases. When x z 0, the slope of the tangent line to x is O. For x > O, f§(x) is positive and increasing. This graph matches (a). f3(x) is an oscillating function like the cosine. Everywhere that the graph of f3 has a horizontal tangent we expect f; to be zero, and (d) matches this condition. The graph matches with (c). (a) f’ is not defined at x = 0, l, 4. At these points, the left-hand and right-hand derivatives do not agree. f(x) — rm) 33. For example, X13157 “2:?” = slope of line joining(—4, 0) and (0,2) 12131121316 H) = slope of linejoining (0, 2) and (17 —2) = —4. Since these values are not equal, f’(0) ; lim0 W does not exist. X 4t (b) , .V f on (—4 6) 4143—4—20 2 4 s s I Left-hand derivative: For h < 0, f(O + h) = f(h) = h? (using y = x2 curve) :> h lin(1)_ w : lim “21:0: lim h: 0; h ~+ 0’ h —> 0‘ Right—hand derivative: For h > 0, f(O + h) : f(h) : h (using y : x curve) :> h lingJ+ w ; lim “—E—Q: lim 1:1; h—vO+ h—VOT Then lim w 75 lim w :> the derivative f’(0) does not exist. h _» 0* h h .. 0+ “ ...
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