Unformatted text preview: Quiz /7 NC /3 Total /10 Math 1A  Quiz 2  Febmzry 8, 2008 NAME
1, Show that there is a root of the equation in the given interval. 2x3+x2+2=0, (— 2, — 1) Letf(x)=2x3+xz+2 thenf(— 2)=— 10<0 and f(~1)=l>0
So, since f (x) is a cantirzuousﬁmction, then by I VT there is a mat in the
interval (— 2, — 1) because f (— 2) <0 and f (~ 1) >0 2. Prove that lim 1 ‘
""”(x+3) =00 We need to show that for every M>0 there is a 5>0 such that 1 4 >M
(x + 3)
So we manipulate the right to make it look like the Ieﬁ . . . which will tell us what 8 needs to be‘ (Remember, it is aurjob here topick 8) l 1 l I
>M—>—>x+3 —3 ‘>x+3
(M). M < ) 2 I I So this tells us we should pick 8 = ~1—1
M?
1
1 “’ A >M
M7 (x + 3) x+3i<5 implies So, then VM>0 x+3< 3. Use the definition ofthe dexivutive to ﬁnd f ’ (a) for
f (x) = 3 — 2x + 4x2 (3—Zx+4xz)(3—2a+4a2) zen—Fa x a x_a _ —2x+2a+4x1—~4a1_. —2(x'a)+4(Xza‘)
_’H“ x—a —H‘ x—a
=xﬁq—2+4(x+a)=—2+4a+a)=—2+8a NOTECARD BONUS QUESTION: (1 point) Find the limit: . 4—v
vlElhl't‘t—V' ...
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 Spring '08
 WILKENING
 Calculus

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