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# 1550 Practice Test One SolutionsExam

Department
Mathematics
Course Code
MATH 1550
Professor
All

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Practice Test # 1, Math 1550, Spring 2012
In general in your test will be used the problems similar to your WebAssign and home-
works.
1. Given a graph of a function fsketch its derivative fbelow.
-2
-1
1
2
3
4
5
-10
10
20
30
40
50
-1
1
2
3
4
5
-40
-20
20
40
2. Determine the type of each given limit and ﬁnd evaluate them, without approxima-
tions or numerical estimations.
a) lim
x0(4x3+ 2 5) = 425, the type is a “number” (or #).
b) lim
x→−10
x+ 10
x2+ 4x60 =1
16 , the type is 0
0.
c) lim
x1+
ex21
sin x=1
sin 1, the type is a “number” (or #).
d) lim
x5
x5
|5x|=1, the type is 0
0.
1

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2e) lim
h→−∞
tan1(x3x) = π
2, the type is f(−∞).
f) lim
t→−3
1t2
t+ 3 =1
4, the type is 0
0.
g) lim
t01
3t1
t(t+ 3)=1
9, the type is ∞ − ∞.
h) lim
s9
s6
s3DNE, the type is #
0
3. Use the Squeeze Theorem to ﬁnd lim
x0(x3cos 1
x).
Answer: For the Squeeze Theorem we need a double inequation, which bounds the
function.
1cos 1
x1 multiply everything by |x3|
−|x3| ≤ |x|3cos 1
x≤ |x3|.
Then using that −|a| ≤ a≤ |a|:
−|x3| ≤ x3cos 1
x≤ |x3|
Now we take limits on the left and on the right, they should agree for the Squeeze
Theorem to work.
lim
x0(−|x3|) = 0 = lim
x0|x3|.
Hence lim
x0x3cos 1
x= 0.
4. Use the Intermediate Value Theorem to show that the equation 2x5+x+ 1 = 0 has
at least one root in the interval (1,0).
Answer: Deﬁne f(x) = 2x5+x+ 1. This is a polynomial function, thus it’s continuous.
Calculating f(1) = 2 and f(0) = 1 we can see that 0 lies between -2 and 1. Then
by Intermediate Value Theorem there is such x=cin the interval (1,0) so f(c) =
2c5+c+ 1 = 0.
5. a ) A ball is dropped from a state of rest at time t= 0. The distance traveled after
tseconds is s(t) = 10 16t2ft. Compute the average velocity over time interval [0.6,0.7].
b) ﬁnd a slope of the secant line of the graph below over the interval [-1.8, -0.8]. What
can you say about slopes of tangent lines at x=-2, -0.8, -0.5, 0.2?

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