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MA 140 - Review for Midterm Key - Spring 2009
- If 2 x 1 f (x) x^2 for 0 < x < 3 , Önd
lim x! 1
f (x)
Since lim x! 1
(2x 1) = 1
and lim x! 1
x 2 = 1
The Squeeze Theorem gives that lim x! 1
f (x) = 1
- Show that the equation 2 x^7 + 4x^3 x^2 3 x + 1 = 0 has at least one real root. Justify your work!
Let f (x) = 2x^7 + 4x^3 x^2 3 x + 1. Then, f (x) is continuous everywhere, since it is a polynomial. Now,
f ( 1) = 2 ( 1)
7
3 ( 1)
2 3 ( 1) + 1
= 3 < 0
and
f (0) = 2 (0)
7
3 (0)
2 3 (0) + 1
= 1 > 0
So, by the Intermediate Value Theorem, there exists a c in the interval ( 1 ; 0) such that f (c) = 0. Hence, the above equation has at least one real root.
- Evaluate:
lim x! 3
p 2 x + 10
p 1 x
x^2 + 7x + 12
lim x! 3
p 2 x + 10
p 1 x
x^2 + 7x + 12
= lim x! 3
p 2 x + 10
p 1 x
(x + 4) (x + 3)
p 2 x + 10 +
p 1 x p 2 x + 10 +
p 1 x
= lim x! 3
(2x + 10) (1 x)
(x + 4) (x + 3)
p 2 x + 10 +
p 1 x
= lim x! 3
3 x + 9
(x + 4) (x + 3)
p 2 x + 10 +
p 1 x
= lim x! 3
3 (x + 3)
(x + 4) (x + 3)
p 2 x + 10 +
p 1 x
= lim x! 3
(x + 4)
p 2 x + 10 +
p 1 x
p 2 ( 3) + 10 +
p 1 ( 3)
- Determine a constant a such that the function
f (x) =
ax 2
- ax + 3 if x 2 3 x + 5a if x > 2
is continuous on the entire real line. (State why the function is continuous everywhere, not just at the ìproblemîpoint.)
Since each piece is a polynomial, f (x) is going to be continuous everywhere except possibly at x = 2. Hence, we need to make sure that
lim x! 2 ^
f (x) = lim x! 2 +^
f (x)
lim x! 2
ax 2
= lim x! 2 +^
(3x + 5a)
4 a 2 a + 3 = 6 + 5a
9 = 3 a
a = 3
So, if a = 3, f (x) will be continuous on the entire real line.
- Evaluate: lim x! 1
x 1
x^2 1
lim x! 1
x 1
x^2 1
= lim x! 1
x + 1
(x 1) (x + 1)
(x + 1) (x 1)
= lim x! 1
x + 1 2
(x 1) (x + 1)
= lim x! 1
x 1
(x 1) (x + 1)
= lim x! 1
x + 1
- The graph of f (x) is given below. Evaluate:
(a) lim x! 3
f (x) = 2
(b) lim x! 1 ^
f (x) =
(c) lim x! 1 +^
f (x) = 1
(d) lim x! 1
f (x) = DN E
(e) lim x! 2 ^
f (x) = 1
(f) lim x! 2 +^
f (x) = 3
(g) lim x! 2
f (x) = DN E
(h) lim x! 3
f (x) = 1
Hence, the function has a essential jump discontinuity at x = 1. At x = 3, we have f (3) = 4, so f (3) is deÖned. Checking the limit, we see
lim x! 3
f (x) = lim x! 3
x 2 5
lim x! 3 +^
f (x) = lim x! 3 +^
(4x 8) = 4
Hence, lim x! 3
f (x) = 4 = f (3)
and the function is continuous at x = 3.
- If 3 x + 2 f (x) x 2 6 x + 4, determine lim x! 2
f (x)
JUSTIFY YOUR ANSWER!!!
Since lim x! 2
( 3 x + 2) = 4
and lim x! 2
x
2 6 x + 4
the Squeeze Theorem says lim x! 2
f (x) = 4
- Evaluate: lim x! 2
x 2 4
x
p 3 x 2
lim x! 2
x^2 4
x
p 3 x 2
= lim x! 2
x^2 4
x
p 3 x 2
x +
p 3 x 2
x +
p 3 x 2
= lim x! 2
x^2 4
x +
p 3 x 2
x^2 (3x 2)
= lim x! 2
(x + 2) (x 2)
x +
p 3 x 2
x^2 3 x + 2
= lim x! 2
(x + 2) (x 2)
x +
p 3 x 2
(x 1) (x 2)
= lim x! 2
(x + 2)
x +
p 3 x 2
x 1
p 3 (2) 2
- Evaluate: lim x! 4
x^2 8 x 9
x^2 9 x + 20
lim x! 4
x^2 8 x 9
x^2 9 x + 20
= lim x! 4
(x 9) (x + 1)
(x 4) (x 5)
As x! 4 , x 9 < 0 ; x + 1 > 0 ; x 4 < 0 ; and x 5 < 0 : Also, x 4 is really close to 0. Therefore,
lim x! 4
(x 9) (x + 1)
(x 4) (x 5)
- Discuss the continuity of f (x) =
x 2 5 ; x 1 4 x 3 ; 1 < x < 2 x + 3; x 2
. At any points where f (x) is discontinuous, state whether
this discontinuity is essential or removable.
Since each piece is a polynomial, f (x) will be continuous everywhere expect possibly where the pieces match up (x = 1 and x = 2). At x = 1 :
- Is f ( 1) deÖned? Yes, f ( 1) = 4
- Does lim x! 1
f (x) exist? We have:
lim x! 1
f (x) = lim x! 1
x 2 5
= 4 and
lim x! 1 +^
f (x) = lim x! 1 +^
(4x 3) = 7
Since limx! 1 f (x) 6 = limx! 1 + f (x) ; we see limx! 1 f (x) does not exist. Therefore, f (x) has an essential discon- tinuity at x = 1.
At x = 2:
- Is f (2) deÖned? Yes, f (2) = 5
- Does lim x! 2
f (x) exist? We have:
lim x! 2 ^
f (x) = lim x! 2 ^
(4x 3) = 5 and
lim x! 2 +
f (x) = lim x! 2 +
(x + 3) = 5
3 : Is lim x! 2
f (x) = f (2)? Yes
So, f (x) is continuous at x = 2. Hence f is continuous on ( 1; 1) [ ( 1 ; 1 ) :
- Evaluate: lim x! 2 +
x 2
x^2 4 x + 4
lim x! 2 +
x 2
x^2 4 x + 4
= lim x! 2 +
(x + 8) (x 2)
(x 2)
2
= lim x! 2 +
x + 8
x 2
As x! 2 , x + 8 > 0 and x + 2 > 0 : Since x 2 is really close to 0, we see:
lim x! 2 +
x + 8
x 2
- Evaluate: lim x! 3
x 3 p x + 1 2
lim x! 3
x 3 p x + 1 2
= lim x! 3
x 3 p x + 1 2
p x + 1 + 2 p x + 1 + 2
= lim x! 3
(x 3)
p x + 1 + 2
(x + 1) 4
= lim x! 3
(x 3)
p x + 1 + 2
(x + 1) 4
= lim x! 3
(x 3)
p x + 1 + 2
x 3
= lim x! 3
p x + 1 + 2
- Discuss the continuity of f (x) =
3 x 2 ; x 2 2 x^2 ; 2 < x 3 4 x + 2; x > 3
Since each piece is a polynomial, f (x) will be continuous everywhere expect possibly where the pieces match up (x = 2
- Find dz dw , if^ z^ = sin^
p w
z = sin
w 1 = 2
dz
dw
= cos
w 1 = 2
w 1 = 2
cos (
p w)
2
p w
- If f (x) = 3 x sin
x 2
, f 0 (x) =?
f 0 (x) = 3 x (ln 3) sin
x 2
x cos
x 2
2 x
= 3 x (ln 3) sin
x 2
x 2
- If j (z) = (sin z)
tan z , d dz [j^ (z)] =^?
y = (sin z)
tan z
ln y = ln (sin z)
tan z
ln y = (tan z) ln (sin z)
1
y
dy
dx
sec 2 z
ln (sin z) + (tan z)
sin z
cos z
dy
dx
= (sin z)
tan z sec 2 z
ln (sin z) + 1
- j(x) = (x + 1)cos^ x. j^0 (x) =
y = (x + 1)
cos x
ln y = ln (x + 1)
cos x
ln y = (cos x) ln (x + 1)
1
y
dy
dx
= ( sin x) ln (x + 1) + (cos x)
x + 1
dy
dx
= (x + 1)
cos x
( sin x) ln (x + 1) +
cos x
x + 1
- y = (sec x)
x^2 , y 0 =?
y = (sec x)
x^2
ln y = ln (sec x)
x^2
ln y = x 2 ln (sec x)
1
y
dy
dx
= 2 x ln (sec x) + x 2
sec x
sec x tan x
dy
dx
= (sec x)
x^2 2 x ln (sec x) + x 2 tan x
- r (k) = ln
q k^3 sin k+cot k ; r 0 (k) =?
r (k) = ln
r k^3
sin k + cot k
= ln
k^3
sin k + cot k
ln
k 3
sin k + cot k
ln k 3 ln (sin k + cot k)
(3 ln k ln (sin k + cot k))
r 0 (k) =
k
sin k + cot k
cos k csc 2 k
k
cos k csc 2 k
sin k + cot k
- Find g 00 (t) if g (t) =
3 p t^4 + 1.
g (t) =
3 p t^4 + 1
t
4
g 0 (t) =
t 4
4 t 3
t 3
t 4
g 00 (t) = 4 t 2
t 4
t 3
t 4
4 t 3
= 4 t 2
t 4
t 6
t 4
- Given the following information: f (1) = 2; f 0 (1) = 5 ; f ( 3) = 8 , f 0 ( 3) = 2, g (x) = f ( 3 x), h (x) = f (x) g(x)
j (x) = f (x) h (x), evaluate: (a) g (1) (b) g 0 (1) (c) h (1) (d) h 0 (1) (e) j (1) (f) j 0 (1)
(a)
g (1) = f ( 3 (1))
= f ( 3) = 8
(b)
g 0 (x) = f 0 ( 3 x) ( 3)
g 0 (1) = f 0 ( 3) ( 3)
= 2 ( 3) = 6
(c)
h (1) =
f (1)
g (1)
(d)
h 0 (x) =
f 0 (x) g (x) f (x) g 0 (x)
[g (x)]
2
h 0 (1) =
f 0 (1) g (1) f (1) g 0 (1)
[g (1)]
2
[ 8]
We need these pieces to match up as well:
lim x! 1
f 0 (x) = lim x! 1 +
f 0 (x)
lim x! 1
3 ax 2
= lim x! 1 +^
2 bx
3 a + b = 2 b
Plugging in 2 for a into this equation, we have:
3 ( 1) + b = 2 b
3 = b
- Suppose j(x) = f ( 3 x), k(x) = g(x)j(x), and l(x) =
g(x) h(x). Suppose^ f; g; h; f^
0 ; g 0 ; and h 0 have the following values:
x -3 -2 0 1 6 f (x) 8 6 4 2 5 g(x) 4 9 3 0 2
h(x) 6 1 -1 -2 - f 0 (x) -2 5 -6 9 1
g 0 (x) -1 -3 4 9 8 h 0 (x) -2 -5 -7 -2 1 2
Find the following: (a) j^0 ( 2) (b) k^0 (1) (c) l^0 (0)
(a)
j 0 (x) = f 0 ( 3 x) ( 3)
j 0 ( 2) = 3 f 0 ( 3 ( 2))
= 3 f 0 (6)
= 3 (1)
= 3
(b)
k 0 (x) = g 0 (x) j (x) + g (x) j 0 (x)
= g 0 (x) f ( 3 x) + g (x) ( 3 f 0 ( 3 x))
k 0 (1) = g 0 (1) f ( 3 (1)) + g (1) ( 3 f 0 ( 3 (1)))
= g 0 (1) f ( 3) + g (1) ( 3 f 0 ( 3))
= 9 (8) + 0 ( 3 ( 2))
= 72
(c)
l 0 (x) =
g^0 (x) h (x) g (x) h^0 (x)
[h (x)]
2
l 0 (0) =
g 0 (0) h (0) g (0) h 0 (0)
[h (0)]
2
[ 1]
2
- g(t) = (tan x)(5x 7) 36 , g 0 (t) =?
g 0 (t) =
sec 2 t
(5t 7)
36
36 (5t 7)
35 5
sec 2 t
(5t 7)
36
35
- k(z) =
csc(2z^2 +3z) z^3 7 ,^
dk dz =?
k 0 (x) =
csc
2 z 2
cot
2 z 2
(4z + 3)
z 3 7
csc
2 z 2
3 z 2
(z^3 7)
2
(4z + 3)
z^3 7
csc
2 z^2 + 3z
cot
2 z^2 + 3z
3 z^2 csc
2 z^2 + 3z
(z^3 7)
2
- Sketch a graph of a function f (x) that satisÖes the following conditions: (a) lim x!
f (x) = 2 (b) f ( 2) is undeÖned (c)
lim x! 2
f (x) = 3 (d) lim x! 2 +
f (x) = 1 (e) f 0 (1) = 0 (f) lim x! 2
f (x) = 1 (g) lim x! 2 +
f (x) = 1 (h) f (x) is continuous at
x = 4, but f 0 (4) does not exist.
There are several (in fact, inÖnitely many) possibilities. Hereís one:
- Evaluate the following: lim x!
x 2 5 x 1 p 3 x^4 + x^2 + 2
The largest power of x that occurs in the denominator is
p x^4 = x 2
So, we divide top and bottom by x 2 to get
lim x!
x^2 5 x 1 p 3 x^4 + x^2 + 2
= lim x!
x^2 5 x 1 x^2 p 3 x^4 +x^2 + x^2
= lim x!
x^2 x^2 ^
5 x x^2 ^
1 x^2 p 3 x^4 +x^2 + x^2
= lim x!
5 x
1 p^ x^2 3 x^4 +x^2 + x^2
(c) The velocity of the ball the instant it hits the ground.
The velocity of the ball will be s^0 (6). So, plugging 6 in for a, we have the velocity as
s 0 (6) = 32 (6) + 80
= 112 m/s
- (a) Using the formal deÖnition of derivative, Önd f 0 (x) if f (x) =
x
. (b) Find an equation of the line tangent to the
graph of f (x) at x = 1
(a)
f 0 (x) = lim h! 0
f (x + h) f (x)
h
= lim h! 0
2 x+h ^
2 x h
= lim h! 0
2 x x(x+h)
2(x+h) x(x+h) h
= lim h! 0
2 x 2(x+h) x(x+h)
h
= lim h! 0
2 h x(x+h) h
= lim h! 0
2 h
x (x + h)
h
= lim h! 0
x (x + h)
x^2
(b) The slope of the tangent line is:
f
0
1 2
The y-coordinate of the point is:
f
1 2 = 4
So, an equation of the tangent line is y + 4 = 8
x + 1 2
- Use the deÖnition of the derivative to Önd f 0 (x) if f (x) = 3x 2 7 x 6.
f 0 (x) = lim h! 0
f (x + h) f (x)
h
= lim h! 0
3 (x + h)
2 7 (x + h) 6
3 x 2 7 x 6
h
= lim h! 0
x^2 + 2xh + h^2
7 (x + h) 6 3 x^2 + 7x + 6
h
= lim h! 0
3 x^2 + 6xh + 3h^2 7 x 7 h 6 3 x^2 + 7x + 6
h
= lim h! 0
6 xh + 3h 2 7 h
h
= lim h! 0
h (6x + 3h 7)
h = lim h! 0
(6x + 3h 7)
= 6 x + 7
- Find all the vertical and horizontal asymptotes of the function f (x) =
2 x 2 x 3
x^2 4 x 5
. Be sure to provide reasons for your
answers. Discuss the behavior of the function around any of these asymptotes.
To determine the vertical asymptotes, we factor that top and bottom of f (x):
f (x) =
2 x 2 x 3
x^2 4 x 5
(2x 3) (x + 1)
(x 5) (x + 1)
2 x 3
x 5
; x 6 = 1
There is no vertical asymptote at x = 1. However, since
lim x! 5
2 x 3
x 5
lim x! 5 +
2 x 3
x 5
There is a vertical asymptote at x = 5.
To determine the horizontal asymptote, we compute:
lim x!
f (x) = lim x!
2 x 3
x 5
= lim x!
3 x 1 5 x = 2
and
lim x!
f (x) = lim x!
2 x 3
x 5
= lim x!
3 x 1 5 x = 2
Hence, the function as a horizontal asymptote of y = 2.
- USING THE DEFINITION OF THE DERIVATIVE, Önd an equation of the tangent line to the graph of
f (x) =
p x 1 at x = 5.
f
0 (x) = lim h! 0
f (x + h) f (x)
h
= lim h! 0
p x + h 1
p x 1
h
= lim h! 0
p x + h 1
p x 1
h
p x + h 1 +
p x 1 p x + h 1 +
p x 1
= lim h! 0
(x + h 1) (x 1)
h
p x + h 1 +
p x 1
= lim h! 0
h
h
p x + h 1 +
p x 1
= lim h! 0
p x + h 1 +
p x 1
p x + 0 1 +
p x 1
p x 1
To Önd the slope of the tangent line, we evaluate
f
0 (5) =
p 5 1
- Suppose y is an implicit function of x. Find
d^2 y dx^2 if^ x
3
x
3
3 = 1
3 x 2
3 y
2 dy dx
= 3 x
2
dy
dx
x 2
y^2
d 2 y
dx^2
2 xy^2 x^2
2 y dy dx
(y^2 )
2
2 xy^2 2 x^2 y
x^2 y^2
y^4
2 xy^3 + 2x^4
y^5
- Suppose y is an implicit function of x. (a) Find dy dx if y = sin(xy). (b) Find the equation of the line tangent to the graph of this equation at the point
5 3
1 2
(a)
y = sin(xy)
dy
dx
= cos (xy)
y + x
dy
dx
dy
dx
= y cos (xy) + x cos (xy)
dy
dx dy
dx
x cos (xy)
dy
dx
= y cos (xy)
dy
dx
(1 x cos (xy)) = y cos (xy)
dy
dx
y cos (xy)
1 x cos (xy)
(b)
dy
dx (^) (x;y)= ( 5 3 ;^
1 2 )
1 2 cos^
5 3 ^
1 2
5 3 cos
5 3
1 2
1 2
p 3 2
5 3
p 3 2
p 3
12 + 10
p 3
So, an equation of the tangent line is:
y
p 3
12 + 10
p 3
x
- Find an equation of the line tangent to the graph of ln (xy + 1) = sin y + tan 1 x at the point (0; 0).
Using implicit di§erentiation:
ln (xy + 1) = sin y + tan 1 x
1
xy + 1
y + x
dy
dx
= cos y
dy
dx
1 + x^2
y
xy + 1
x
xy + 1
dy
dx
= cos y
dy
dx
1 + x^2
x
xy + 1
dy
dx
cos y
dy
dx
1 + x^2
y
xy + 1
dy
dx
x
xy + 1
cos y
1 + x^2
y
xy + 1
dy
dx
1 1+x^2 ^
y xy+ x xy+ cos y
The slope of the tangent line is:
dy
dx (^) (x;y)=(0;0)
1 1+0^2
0 0(0)+ 0 0(0)+1 ^ cos 0
Hence, an equation of the tangent line is y 0 = 1 (x 0) ; or y = x:
- Find
d 2 y
dx^2
if x 3
x
3
3 6 xy = 0
3 x 2
6 y + 6x
dy
dx
3 x 2
6 y 6 x
dy
dx
3 y 2 dy dx
6 x
dy
dx
= 6 y 3 x 2
dy
dx
3 y 2 6 x
= 6 y 3 x 2
dy
dx
6 y 3 x 2
3 y^2 6 x
d 2 y
dx^2
dy dx ^6 x
3 y 2 6 x
6 y 3 x 2
6 y
dy dx ^6
(3y^2 6 x)
2
6 y 3 x^2 3 y^2 6 x
6 x
3 y 2 6 x
6 y 3 x 2
6 y
6 y 3 x^2 3 y^2 6 x
(3y^2 6 x)
2
6 y 3 x^2
6 x
3 y^2 6 x
3 y^2 6 x
6 y 3 x^2
6 y
6 y 3 x^2
3 y^2 6 x
(3y^2 6 x)
3
- A hot-air balloon rising straight up from a level Öeld is tracked by a range Önder 500 feet from the lifto§ point. At
the moment the range Öndersíelevation angle is 4 ; the angle is increasing at a rate of 0.14 rad/min. How fast is the balloon rising at that moment?
θ
500
h
