Solutions - Pat Rossi

MTH 1125 - Test #1 - Solutions
Summer 2006
Pat Rossi
Name
Instructions. Show CLEARLY how you arrive at your answers.
1. Compute: limx→2
x2 −x−2
x−2
=
1. Try plugging in:
limx→2
x2 −x−2
x−2
=
(2)2 −(2)−2
(2)−2
0
0
=
No good Zero Divide!
2. Try factoring and cancelling:
limx→2
x2 −x−2
x−2
i.e., limx→2
2. Compute: limx→3
x2 +5
x2 −3
(x−2)(x+1)
x−2
= limx→2
x2 −x−2
x−2
= limx→2 (x + 1) = (2) + 1 = 3
=3
=
1. Try plugging in:
limx→3
x2 +5
x2 −3
=
i.e., limx→3
3. Compute: limx→5
(3)2 +5
(3)2 −3
x2 +5
x2 −3
x2 +5
x2 −3x−10
=
=
14
6
=
7
3
7
3
=
1. Try plugging in:
limx→5
x2 +5
x2 −3x−10
=
(5)2 +5
(5)2 −3(5)−10
=
30
0
No good Zero Divide!
2. Try factoring and cancelling:
No good Step 2 only works when Step 1 yields
0
0
3. Examine one-sided limits
limx→5−
x2 +5
x2 −3x−10
= limx→5−
x2 +5
(x+2)(x−5)
=
30
(7)(−ε)
= limx→5+
x2 +5
(x+2)(x−5)
=
30
(7)(ε)
=
( 307 )
(−ε)
= −∞
x → 5−
⇒ x<5
⇒ x−5<0
limx→5+
x2 +5
x2 −3x−10
=
( 307 )
(+ε)
= +∞
x → 5+
⇒ x>5
⇒ x−5>0
Since the one sided limits are not equal, limx→5
x2 +5
x2 −3x−10
does not exist.
4. f (x) = 3x2 + 2x − 5. Compute f 0 (x) using the definition of derivative (i.e., using the
“limiting process”).
f 0 (x) = lim∆x→0
f (x+∆x)−f (x)
∆x
= lim∆x→0
(3(x+∆x)2 +2(x+∆x)−5)−(3x2 +2x−5)
∆x
= lim∆x→0
(3(x2 +2x∆x+∆x2 )+2(x+∆x)−5)−(3x2 +2x−5)
= lim∆x→0
(3(x2 +2x∆x+∆x2 )+2(x+∆x)−5)−(3x2 +2x−5)
= lim∆x→0
3x2 +6x∆x+3∆x2 +2x+2∆x−5−3x2 −2x+5
∆x
= lim∆x→0
∆x(6x+3∆x+2)
∆x
∆x
∆x
= lim∆x→0
6x∆x+3∆x2 +2∆x
∆x
= lim∆x→0 (6x + 3∆x + 2) = 6x + 3 (0) + 2 = 6x + 2
i.e., f 0 (x) = 6x + 2
5. f (x) =
4x+5
.
2x−2
Find all vertical and horizontal asymptotes. Graph f (x) .
Verticals Find x-values that cause division by zero.
⇒ 2x − 2 = 0
⇒ 2x = 2
⇒ x = 1 (possible vertical asymptote)
2
Examine the one-sided limits at x = 1
limx→1−
4x+5
2x−2
= limx→1−
( 92 )
4x+5
(2)(x−1)
=
9
(2)(−ε)
=
(−ε)
4x+5
(2)(x−1)
=
9
(2)(+ε)
=
(+ε)
= −∞
x → 1−
⇒ x<1
⇒ x−1<0
limx→1+
4x+5
2x−2
= limx→1+
( 92 )
= +∞
x → 1+
⇒ x>1
⇒ x−1>0
Since the one-sided limits are infinite, x = 1 is a vertical asymptote.
Horizontals Let x → ±∞
limx→−∞
4x+5
2x−2
= limx→−∞ 4x
= limx→−∞ 2 = 2
2x
limx→+∞
4x+5
2x−2
= limx→+∞ 4x
= limx→+∞ 2 = 2
2x
Finite limits indicate that y = 2 is a horizontal asymptote.
Graph:
y=2
x=1
3
6. Compute:
d
dx
Rewrite:
d
dx
d
dx
h
[5x4 + 3x3 + 8x2 + 3x − 7 +
1
d
dx
i
1
5x4 + 3x3 + 8x2 + 3x − 7 + x 2 = 20x3 + 9x2 + 16x + 3 + 12 x− 2
[5x4 + 3x3 + 8x2 + 3x − 7 +
7. Compute:
√
x] =
√
1
x] = 20x3 + 9x2 + 16x + 3 + 12 x− 2
[(3x4 + 2x2 + 5) (8x5 − 2x3 )] =
Use the product rule:
⎡
⎢³ 4
d ⎢
3x
dx ⎣
|
i.e.,
d
dx
⎤
´³
´⎥
+ 2x2 + 5 8x5 − 2x3 ⎥
⎦
{z
}|
{z
}
1st
2n d
d
dx
h
sin(x)
(5x2 +2x+2)
Use the quotient rule:
⎡
⎤
to p
z }| {
⎢
⎥
⎢
sin (x) ´ ⎥
⎥
d ⎢³
⎢
⎥
dx ⎢
2
⎥
⎣ 5x + 2x + 2 ⎦
{z
}
|
i
´
{z
|
1st prime
}|
{z
2n d
}
=
to p p rim e
z
z }| {³
h
sin(x)
(5x2 +2x+2)
Extra Credit Compute: limx→2
i
=
b o tto m
}|
b o tto m p rim e
to p
´{ z }| {z }| {
cos (x) 5x2 + 2x + 2
=
³
|
b o tto m
d
dx
´³
³
|
{z
2n d prime
´³
}|
´
3x4 + 2x2 + 5
{z
1st
}
[(3x4 + 2x2 + 5) (8x5 − 2x3 )] = (12x3 + 4x) (8x5 − 2x3 ) + (40x4 − 6x2 ) (3x4 + 2x2 + 5)
8. Compute:
i.e.,
³
= 12x3 + 4x 8x5 − 2x3 + 40x4 − 6x2
−(10x
´2
5x2 + 2x + 2
{z
b o tto m sq u a red
+ 2)sin (x)
}
cos(x)(5x2 +2x+2)−(10x+2) sin(x)
(5x2 +2x+2)2
√
x+7−3
x−2
=
1. Try plugging in:
√
limx→2 x+7−3
x−2
=
√
(2)+7−3
(2)−2
=
No good Zero Divide!
0
0
4
2. Try factoring and cancelling:
√
limx→2 x+7−3
x−2
=
√
limx→2 x+7−3
·
x−2
= limx→2
= limx→2
i.e., limx→2
√
x+7−3
x−2
=
√
√
2
x+7+3
( x+7) −(3)2
√
= limx→2 (x−2) √x+7+3
[
]
x+7+3
|
(x+7)−9
√
(x−2)[ x+7+3]
[
1
√
x+7+3]
1
6
5
{z
}
=1
= limx→2
= h√
1
i
(2)+7+3
(x−2)
√
(x−2)[ x+7+3]
=
1
6