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Franklin and Marshall College
Department of Mathematics
MAT 109
Exam 1
Fall 2014
Name: SOLUTIONS
Instructions:
• You have 65 minutes to take this exam.
• You are allowed only to have a writing implement (pen, pencil, eraser) while taking this exam.
• No internet, notes, or any other outside aids are permitted. You may use the backs of pages
for scratchwork.
• Simplify your answers completely. For example, expressions such as
√
4/9 + 5/3, (1 − 32 )/(−15),
16, and ln(e2 )
should be simplified further, whereas expressions like
√
√
ln(3)/15 + 1/15, (ln(3) + 1)/15,
12, 7/ 2,
and
ln(5)
are considered to be fully simplified.
• Please read each question carefully. Show ALL work clearly in the space provided.
• If necessary, please use the symbols ∞, −∞, and DNE, with their intended connotation.
• Full credit will be awarded to solutions that are complete, legible, logically presented and
notationally sound.
• The point value of each question is indicated after its statement.
Grading - For Administrative Use Only
Question:
1
2
3
4
Total
Points:
30
20
20
25
95
Score:
MAT 109, Manack
Exam 1
1. Let f be the function
[30]
10x4 − 70x3 + 100x2
f (x) =
x(x − 5)(x + 5)2
(a) Find the domain of f . Express your answer in interval notation.
(−∞, −5) ∪ (−5, 0) ∪ (0, 5) ∪ (5, ∞)
(b) Evaluate the following limits. If a limit does not exist, you must justify your rationale
(with words):
Factor numerator
10x2 (x − 2)(x − 5)
f (x) =
x(x − 5)(x + 5)2
(a) lim f (x) = 0 (Factor, cancel, then DSP)
x→0
(b) lim f (x) =
x→5
10 ∗ (52 )3
3
= (Factor, cancel, then DSP)
2
5 ∗ 10
2
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Math 109, Manack
Exam 1
For convenience, here is the function f (x) again:
f (x) =
(c) lim f (x) = lim
x→−5
x→−5
10x4 − 70x3 + 100x2
x(x − 5)(x + 5)2
10x(x − 2)
= ∞ (little h method)
(x + 5)2
(d) Write down the vertical asymptotes of f : x = −5
10 − 70/x + 100/x2
= 10/1 = 10
x→∞ (1 − 5/x)(1 + 5/x)2
(e) lim f (x) = lim
x→∞
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Math 109, Manack
Exam 1
2. Below is the graph of g(x):
[20]
Use the graph of g(x) to answer each request. No justification needed.
(a) Find the domain of g. Write your answer in interval notation.
(−∞, −5) ∪ (−5, −3) ∪ (−3, 9) ∪ (9, ∞)
(b) g(2) = −2
(c) lim g(x) = 3
x→2+
(d) lim g(x) DNE
x→2
(e) lim 3g(x) − g(x)2 = 3(3) − (3)2 = 0
x→6
(f) lim g(x) = ∞
x→−3
(g)
lim g(x) = ∞
x→−5−
(h) lim g(x) = 3
x→∞
(i)
lim g(x) = 1
x→−∞
(j) Write down all x-values where g is discontinuous. Then classify each discontinuity as
removable, jump (with asymptotic as a special case), or essential.
x = 9 removable, x = 6 removable, x = 2 jump, x = −3 jump (asymptotic), x = −5 jump
(asymptotic).
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Math 109, Manack
Exam 1
3. Let f (x) = 1 − 5e3x .
[20]
(a) Starting with the function g(x) = e3x , list the graph transformations, in the correct order,
which take the graph of g(x) to the graph of f (x).
reflect across x axis, then
vertically stretch by 5 units, then
shift up by 1 units.
(b) Determine the inverse function f −1 (x).
y = 1 − 5e3x
x = 1 − 5e3y
x−1
= e3y
−5
x−1
ln(
) = 3y
−5
1
x−1
ln(
)=y
3
−5
1
x−1
f −1 (x) = ln(
)
3
−5
(c) Let h(x) = x2 − x. Compute (h ◦ f )(x). Simplify your answer completely.
(h ◦ f )(x) = h(f (x)) = h(1 − 5e3x )
= (1 − 5e3x )2 − (1 − 5e3x ) = 1 − 10e3x + 25e9x − 1 + 5e3x = −5e3x + 25e6x
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Math 109, Manack
Exam 1
4. Evaluate the following limits. If a limit does not exist, you must justify your rationale (with
words):
(a) lim ln(5x2 + 4) − 2 ln(5 + 3x) = lim ln(5x2 + 4) − ln((5 + 3x)2 )
x→∞
x→∞
= lim ln((5x2 + 4)/(5 + 3x)2 ) = lim ln((5x2 + 4)/(25 + 30x + 9x2 )) = ln 5/9.
x→∞
x→∞
√
√ 2
2
4
4
4
(b) lim
√ (cos(x + π/6)) = (cos( π + π/6)) = (−( 3/2)) = 9/16 by DSP.
x→ π
|x| + 2x
x→0 |x| − 2x
(Hint: |x| is piecewise function, remember?)
(c) lim
lim
x→0+
lim
x→0−
x + 2x
|x| + 2x
= lim
= lim −3 = −3
|x| − 2x x→0+ x − 2x x→0+
|x| + 2x
−x + 2x
= lim
= lim −1/3 = −1/3
−
|x| − 2x x→0 −x − 2x x→0−
|x| + 2x
DNE
x→0 |x| − 2x
By one sided criterion, lim
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