MAT 137Y, 2005–2006 Solutions to Term Test 2
1.
(10%) (a) Find a constant kfor which
lim
x→0
sinx−x−kx3
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
x→0
cosx−1−3kx2
5x4=lim
x→0−sinx−6kx
20x3=lim
x→0−cosx−6k
60x2.
The resulting limit does not exist unless the numerator goes to zero, which occurs when k=−1
6.
This gives
L=lim
x→0−cosx+1
60x2
which by L’Hˆ
opital’s Rule gives
L=lim
x→0
sinx
120x=1
120 lim
x→0
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
a−1
4
=2a.
Solving for a,
2a(a−1
4) = a2+3
2=⇒2a2−a
2=a2+3
2
=⇒a2−a
2−3
2=0=⇒2a2−a−3=0
=⇒(2a−3)(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=3x−1
4=⇒y=3x−9
4,
y+3
2=−2x−1
4=⇒y=−2x−2.
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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
(x−2)(y−3)=(x−2)180
x−3=186 −3x−360
x=A(x),x∈(2,60).
Differentiating,
A0(x) = −3+360
x2=0=⇒x2=120 =⇒x=2√30.
This gives an absolute maximum since A0(x)>0 for 0 <x<2√30 and A0(x)<0 when x>2√30.
(Alternatively, lim
x→2+A(x) = lim
x→60−A(x) = 0, so the critical point must yield an absolute maximum.)
When x=2√30 then y=180/(2√30)so the dimensions are 2√30 ×90
√30 or 2√30 ×3√30.
3. Consider the function
f(x) = x
x2+9.
(4%) (a) Show that f00(x) = 2x(x2−27)
(x2+9)3.
f0(x) = (x2+9)−x(2x)
(x2+9)2=9−x2
(x2+9)2
=⇒f00(x) = (−2x)(x2+9)2−(9−x2)·2(x2+9)(2x)
(x2+9)4=(−2x)(x2+9)−4x(9−x2)
(x2+9)3
=2x3−54x
(x2+9)3=2x(x2−27)
(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) = 9−x2
(x2+9)2=0 when x=±3, which are the critical points for f.
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