Unformatted text preview: SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS 349 26.f(:c)=ac3+cm=a:(;c2+c) => f/(m)=3m2+c : f”(x)=6:c
20 20 _50"5_5 5 5 5
ﬁll—0.5 _20 —20 Cziﬁ c:0 (3:6 arc—intercepts: When c 2 0, 0 is the only sic—intercept. When 0 < (J, the m—intercepts are 0 and ::\/——c.
yintercept = f(0) : 0. f is odd. so the graph is symmetric with respect to the origin. f"(a:) < 0 for :c < 0 and
f”(m) > O for ac > 0, so f is CD on ('00. 0) and CU on (0. oo) . The origin is the only inﬂection point. Ifc > 0. then f’(:r) > 0 for all m. so f is increasing and has no local maximum or minimum. If c = 0. then f’(:c) 2 O with equality at :1: : 0. so again f is increasing and has no local maximum or minimum. Ifc<0.thenf’(m):3[:c2—( (—)]c/3 —3(m+ —C/3><:c— —c/3).
sof'(at)>00n (—oo.— —c/3) and< —c/3.oo);
( JT/a/T/ei) Itfollowsthatf(— —c/3):*§c —c/3is
c\/T/3isaiocai minimum value. As 0 decreases (toward more negative values), the local a local maximum value and f( —c/3) = calm maximum and minimum move further apart. There is no absolute maximum or minimum value. The only L z / ‘20
. . . . L‘ : 4
transrtional value of c corresponding to a change in character of the graph L. : 6
c : 8
is c = 0. 27. f(:c) = m4 + cm2 2 :32 (m2 + 0). Note that f is an even function. For c 2 0. the only m—intercept is the point (0. 0).
We calculate f’(:c) : 4353 + 2cm : 4w(w2 + gs) => f”(:c) : 12332 + 2c. Ifc 3 0. m : 0 is the only critical point and there is no inﬂection point. As we can see from the examples. there is no Change in the basic shape of the graph for c 2 0; it merely becomes steeper as 0 increases. For c : 0, the graph is the simple curve y = $4. For c < 0, there are ac—intercepts at 0 and at ::\/—c. Also. there is a maximum at (0, 0), and there are minima at (i1/——c ‘ZC 2). As c —> ‘00. the m—coordinates of these minima get larger in absolute value, and the minimum points move downward. There are inﬂection points at (:: —%c — 3—566 2.) which also move away from the origin as c —> —oo. ...
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 Spring '10
 Ban
 Calculus

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