2005 Test 2 solution

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10 Apr 2012
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Department
Course
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MAT 137Y, 2005–2006 Solutions to Term Test 2
1.
(10%) (a) Find a constant kfor which
lim
x0
sinxxkx3
x5
exists and the limit is also non-zero, and determine the value of the limit.
Let Lbe the limit above. Regardless of the value of k, the limit is of the form (0
0), so by applying
L’Hˆ
opital’s Rule three times,
L=lim
x0
cosx13kx2
5x4=lim
x0sinx6kx
20x3=lim
x0cosx6k
60x2.
The resulting limit does not exist unless the numerator goes to zero, which occurs when k=1
6.
This gives
L=lim
x0cosx+1
60x2
which by L’Hˆ
opital’s Rule gives
L=lim
x0
sinx
120x=1
120 lim
x0
sinx
x=1
120.
(10%) (b) Find the equation of all tangent lines to the curve y=x2that intersect the point (1
4,3
2).
Draw a picture. Clearly (1
4,3
2)is NOT on the curve.
Suppose the tangent line intersects the curve at the point (a,a2). Using two different ways to
obtain the slope of the tangent line, we have
m=a2+3
2
a1
4
=2a.
Solving for a,
2a(a1
4) = a2+3
2=2a2a
2=a2+3
2
=a2a
23
2=0=2a2a3=0
=(2a3)(a+1) = 0=a=3
2,1.
The corresponding slopes are m=2a=3,2.
Therefore the equations of the tangent lines are
y+3
2=3x1
4=y=3x9
4,
y+3
2=2x1
4=y=2x2.
1
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(12%) 2. A poster is to have an area of 180 square inches with 1 inch margins at the bottom and sides and a 2
inch margin at the top. What dimensions will give the largest printed area? Make sure to verify that
your answer yields a maximum.
Let xbe the width of the poster and ybe the height of the poster.
Since xy =180, it follows that y=180/x.
The printed area is
(x2)(y3)=(x2)180
x3=186 3x360
x=A(x),x(2,60).
Differentiating,
A0(x) = 3+360
x2=0=x2=120 =x=230.
This gives an absolute maximum since A0(x)>0 for 0 <x<230 and A0(x)<0 when x>230.
(Alternatively, lim
x2+A(x) = lim
x60A(x) = 0, so the critical point must yield an absolute maximum.)
When x=230 then y=180/(230)so the dimensions are 230 ×90
30 or 230 ×330.
3. Consider the function
f(x) = x
x2+9.
(4%) (a) Show that f00(x) = 2x(x227)
(x2+9)3.
f0(x) = (x2+9)x(2x)
(x2+9)2=9x2
(x2+9)2
=f00(x) = (2x)(x2+9)2(9x2)·2(x2+9)(2x)
(x2+9)4=(2x)(x2+9)4x(9x2)
(x2+9)3
=2x354x
(x2+9)3=2x(x227)
(x2+9)3.
(4%) (b) Find the domain, all intercepts, and asymptotes.
The domain is R. The only intercept is (0,0).
Since lim
x→±
x
x2+9=0, the horizontal asymptote is y=0.
Since the domain is Rthere are no vertical asymptotes.
(6%) (c) Locate all critical points of f, determine and clearly indicate the intervals for which fis increas-
ing or decreasing, and classify all critical points are local maxima, local minima, cusps, vertical
tangents, or none of the above.
Since f0(x) = 9x2
(x2+9)2=0 when x=±3, which are the critical points for f.
2
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