Unformatted text preview: Quiz/7 NC [3 Total In) Math 1A — Quiz 3 — February 15, 2008 NAME
1‘ The graph of f is given 2
Skewh the graph of f' . 1.5 I " f is shown at right, please ”5
see your quiz for the original
graph off 4.1.5 05 I a 56
.I \5 2, Find the ﬁrst and second derivatives of g (t) = 29051 — Mint
For what values ofn does g” (t) = g (t) 7 g ’ (t) = — 2sint— most
g ” (t) = — Zemt + mint Now for the second question, we must ﬁnd what derivative gives
us back the original ﬁmction, g (t)  So, g ‘” (t) =2sl'nt + most g””(t) =g’y (t) =2cost— mint =g (t) Thus we found that when n = 4 we gEt g (t) back  Additionally, this is also true if n is any multiple of 4 t 3‘ For what values ofx does the graph ofﬂx) = 2x3 + 151:“ + 36x + 6 have a
horizontal tangent? We muslfind where f (x) has horizontal tangents . . . or where the derivative
iszero ~ So,
f’(x)=6xZ +30x +36 So, f(x)=0 —> 6x2 +30x+36 =0 —»6(x‘+5x+6)=0 —+ (x+3)(x+2)=0 —)x=——3 orx=—2
Thusf(x) has han‘zontaltangents when x =— 3 or): =~ 2 NOTECARD BONUS QUESTION: (1 point) Find the equations of the tangent line and the normal line to the curve at the given
point y=6cosx (g , 3) In order toﬁnd the equations of the tangent and normal lines, we must
find the slopes of each of these lines  So, we must ﬁrst ﬁnd the derivative
of the function  y ’ = — 6sinx Now thederivative ofyatg isy’(g)=i 65in (:)=—6(‘L2:)='— 3 «/§ So the tangent line to theﬁmctton at (% , 3) has slope — 3 \/3_ Thus the equation of the tangent line is: y — 3 =— 3 V? (x — 135) Now we know that the tangent line and normal line are perpendicular.
thus their slopes must be negative reciprocals ‘ So the normal line to the ﬂmction at (g . 3) has slope 3 \1/3.
Thus the equation afthe normal line is: y — 3 = 3 3/? ( — g) ...
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 Spring '08
 WILKENING
 Calculus

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