Midterm 1 Solutions
Monday, 10/24/2011
1. (10 points)
Consider the function
y = f (x) =
ex
.
1 + 2ex
(a) (2 points) What is the domain of f ? Express your answer using
interval notation.
Solution:
We must exclude the possibility that 1+2ex = 0.
But this would mean ex = − 21 , which is impossible since ex > 0
for all x. So the domain is (−∞, ∞).
(b) (5 points)
Solution:
Compute f −1 , the inverse of f .
We solve for x in the formula for f above.
ex
y=
1 + 2ex
x
(1 + 2e )y = ex
y + 2yex = ex
y = ex − 2yex
y = (1 − 2y)ex
y
ex =
1 − 2y
y
x = ln
1 − 2y
x
So the inverse function is f −1 (x) = ln 1−2x
.
(c) (3 points)
What is the range of f ? Express your answer using
interval notation.
Solution:
The easy way to do this problem is to remember
that the range of f is the domain of f −1 . So here we must have
x
>0
1 − 2x
in order for the natural log to make sense. Drawing a number line
one may check that this occurs for (0, 12 ).
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2. (10 points) Determine the following limits of the function f (x) whose
graph is given above. For infinite limits, use ±∞ when appropriate. If
no limit exists, write Does Not Exist or (DNE).
(a) lim f (x) = 0
x→∞
(b) lim f (x) = 2
x→−8
(c) lim f (x) = DN E
x→2
(d) lim f (x) = DN E
x→4
(e) lim f (x) = DN E
x→−4
(f) lim f (x) = +∞
x→−2
(g)
lim
x→(−6)+
f (x) = 1
(h) lim− f (x) = −∞
x→4
(i)
lim
x→(−4)+
f (x) = 3
(j) lim f (x) = +∞
x→−∞
3. (35 points)
Compute the limit algebraically. You must show your
work at a level appropriate for the particular problem at hand. Do not
use L’Hopital’s rule, if you know it (definitely do not use it if you don’t
know it!). If a limit does not exist, show why.
(a)
22 − 1
x2 − 1
=
= 3.
x→2 x − 1
2−1
lim
(b)
x3 + x2 − x − 1
.
x→−1
x+1
Solution:
This limit is of the type 0/0. Since the function
is a rational function, we look for a cancellation of zeros. Using
polynomial division to write x3 + x2 − x − 1 = (x + 1)(x2 − 1) =
(x + 1)2 (x − 1), we see for x 6= −1,
lim
x3 + x2 − x − 1
(x + 1)2 (x − 1)
=
= (x + 1)(x − 1),
x+1
x+1
2
and so
x3 + x2 − x − 1
lim
= lim (x − 1)(x + 1) = 0.
x→−1
x→−1
x+1
(c)
1 − 2t2 − t4
.
t→−∞ 6 + t − 3t4
Solution:
Here this limit is of the form ∞/∞, so we divide
by the highest power of t in the denominator and simplify:
lim
1−2t2 −t4
t4
6+t−3t4
t4
1
2
4 − t2 −
lim t6
t→−∞ 4 + 13 −
t
t
1 − 2t2 − t4
= lim
lim
t→−∞
t→−∞ 6 + t − 3t4
=
1
3
1
= .
3
(d)
lim ln(sin θ).
θ→π −
Solution:
Here, observe that limθ→π− sin(θ) = 0+ , i.e. the
limit is zero but values approach from the positive side. Since
ln(x) is continuous for x > 0, setting t = sin(θ) we find that
lim ln(sin θ) = lim+ ln(t) = −∞.
θ→π −
t→0
(e)
√
49 + x − 7
.
x→0
x
Solution:
Plugging in, we again get a 0/0 type limit, so we
need to cancel somehow. The top has a square root we want to get
rid of, so we multiply both the top and bottom by the conjugate.
lim
3
√
lim
x→0
√
√
49 + x − 7
( 49 + x − 7)( 49 + x + 7)
√
= lim
x→0
x
x( 49 + x + 7)
(49 + x) − 49
= lim √
x→0 x( 49 + x + 7)
x
= lim √
x→0 x( 49 + x + 7)
1
= lim √
x→0 ( 49 + x + 7)
1
= √
( 49 + 0 + 7)
1
=
14
(f)
2
lim x cos
π .
x
This is a squeeze theorem problem. Since
π −1 ≤ cos
≤1
x
x→0
Solution:
for all x 6= 0, we have, upon multiplication through by x2 ,
π 2
2
−x ≤ x cos
≤ x2 .
x
Since limx→0 −x2 = 0 = limx→0 x2 , we have
π 2
lim x cos
= 0,
x→0
x
by the squeeze theorem.
(g)
lim f (t),
t→1
where
4
 2
t<1
 t,
2,
t=1
f (t) =

cos(πt) + 2 log(t), t > 1
Solution:
Here you must compute left and right hand
limits, since the algebraic definition of f changes exactly at x = 1.
lim f (t) = lim t2 = 1,
t→1−
t→1
and
lim f (t) = lim cos(πt) + 2 log(t) = cos π + 2 log 1 = −1.
t→1
t→1+
So the limt→1 f (t) does not exist.
4. (10 points)
(a) (5 points)
Assume that f and g are continuous functions with
the following limits:
lim f (x) = 5
lim f (x) = 3
x→0
x→5+
lim g(x) = 2
lim g(x) = 5
x→0−
x→5
Calculate the limit
lim g (f (x)).
x→5−
Solution:
Since g and f are continuous, g ◦ f is also
continuous (section 2.4, theorem 9) and so we can calculate limits (including left and right limits) by evaluation. In particular, lim− g (f (x)) = lim g (f (x)) g (f (5)). Since f is continuous,
x→5
x→5
f (5) = lim+ f (x) = 3, and g(f (5)) = g(3). Since g is continuous
x→5
g(3) = lim g(x) = 4 .
x→3
5
lim f (x) = 6
x→2+
lim g(x) = 4
x→3
(b) (5 points)
what is
Using the functions f and g from the question above,
lim ef (x)−g(x) + sin(2πf (x)) + 3f (x)g(x) + ln(16f (x) + g(x))
x→0
Solution:
Again, since f and g are continuous, as are ex ,
sin(x) and ln(x), the function ef (x)−g(x) +sin(2πf (x))+3f (x)g(x)+
ln(16f (x) + g(x) is also continuous, and so the limit can be calculated by direct evaluation. Thus
lim ef (x)−g(x) + sin(2πf (x)) + 3f (x)g(x) + ln(16f (x) + g(x))
x→0
= ef (0)−g(0) + sin(2πf (0)) + 3f (0)g(0) + ln(16f (0) + g(0))
Since f and g are continuous, f (0) = lim f (x) = 5 and g(0) =
x→0
lim g(x) = 5. Plugging this into the above equation gives
x→0
e5−5 + sin(2π5) + 3 · 5 · 5 + ln(16 · 5 + 5)
= e0 + sin(10π) + 75 + ln(85) = 76 + ln(85).
5. (10 points)
This question is about the Intermediate value theorem.
(a) (2 points)
State the Intermediate value theorem precisely.
Solution:
Suppose that f is a continuous function on the
interval [a, b] and let N be any number in between f (a) and f (b)
(with f (a) 6= f (b)). Then there exists a number c in the (open)
interval (a, b) with f (c) = N .
(b) (3 points)
Prove there is a solution to the equation
x4 + x − 3 = 0 on [1, 2].
Solution:
Let P be the polynomial P (x) = x4 +x−3. Since
P is a polynomial, it is continuous; P (1) = −1 and P (2) = 15,
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and so by the Intermediate Value Theorem, since 0 is in between
P (1) and P (2), there exists a number c in (1, 2) with P (c) = 0.
This is the desired solution to the equation.
(c) (2 points)
Give an explicit example of a quadratic polynomial
that has no real root (i.e. state the equation) .
Solution:
The polynomial x2 + 1 has no real roots.
(d) (3 points)
By considering the limit as x → ±∞, explain why
every a cubic polynomial of the form
p(x) = x3 + bx2 + cx + d,
(here b, c, d are real constants) must have at least one real root.
Solution:
The function p(x) is a cubic polynomial and
the x3 dominates at ±∞, so lim p(x) = −∞ and lim p(x) =
x→−∞
x→∞
∞ (similar to x3 ). So, p(x) grows without bound and there is
some (large, negative number) a with p(a) < 0 and some (large,
positive number) b with p(b) > 0. Since p(x) is a polynomial it
is continuous and so by the Intermediate Value Theorem, there is
some number c in (a, b) with p(c) = 0. Thus p has a root.
6. (5 points)
 2
x +2x+1

x < −1
 x2 −1
2
f (x) = ax + 3x + 1 −1 ≤ x < 2


b cos(πx)
2≤x
What are the values, if any, of a and b which make f (x) a continuous
function on (−∞, ∞)?
Solution:
Since each constituent function is continuous on its
domain, we only need to check continuity at the endpoints of the subdomains - i.e., at −1 and 2. To check continuity, we need to see that
lim f (x) = f (l) and in particular, this means that we need to show
x→l
•
lim f (x) = f (−1), and so
x→−1−
lim −
x→−1
x2 + 2x + 1
= a(−1)2 + 3(−1) + 1
x2 − 1
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On the left we have: lim −
x→−1
(x + 1)(x + 1)
(x + 1)(x + 1)
= lim −
=
(x + 1)(x − 1) x→−1 (x + 1)(x − 1)
−1 + 1
(x + 1)
=
=0
x→−1 (x − 1)
−1 − 1
and on the right
a(−1)2 + 3(−1) + 1 = a − 2. Thus a = 2.
lim −
• lim− f (x) = f (2). On the left we have
x→2
lim 2x2 + 3x + 1 = 2 · 22 + 3 · 2 + 1 = 15, and on the right
x→2−
f (2) = b cos(π2) = b. And thus b = 15.
Thus for a = 2 and b = 15, lim f (x) = f (l) for l = −1 and l = 2 and
x→l
thus f is continuous.
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7. (10 points)
Determine if the following statements are true or false.
No justification is needed.
f (x)
does not exist.
x→a
x→a g(x)
Solution:
FALSE. Basically take any limit of the form 0/0
as an example. If you want something concrete, look at problem
3b of this exam.
√
(b) (2 points)
g(x) = 3x2 + 9x + sin(2) is a polynomial.
Solution:
TRUE. Here the
√ exponents on the powers of x
are all nonnegative integers, and 3 and sin(2) are constant real
numbers.
(a) (2 points)
If lim g(x) = 0, then lim
(c) (2 points)
If both lim f (x) and lim g(x) does not exist, then
x→a
x→a
lim (f (x) + g(x)) also does not exist.
x→a
Solution:
FALSE. Take your favourite limit that does
not exist. Say sin(1/x) as x → 0+ . Let f (x) = sin(1/x) and
g(x) = − sin(1/x). Then neither limx→0+ f (x) or limx→0+ g(x)
exists, but lim+ (f (x) + g(x)) = lim 0 = 0
x→0
x→0
(d) (2 points)
If p is a polynomial and b is any real number, then
lim p(x) = p(b).
x→b
Solution:
TRUE. This just says that a polynomial is continuous at any real number b.
(e) (2 points) A function can never cross its horizontal asymptotes.
Solution:
FALSE. A curve can cross its horizontal asymptotes even infinitely many times. One example is the function
1
sin(x) considered on the interval (1, ∞). Here the squeeze thex2
orem shows the amplitudes decrease to zero as x → ∞, and so
y = 0 is an asymptote.
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