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 lefthand and righthand 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
Lefthand 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+ “ ...
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 Spring '10
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