MATH 184 Quiz #2
October 8
Grade:
First Name:
Last Name:
Student-No:
Section:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes.
Marking scheme: 1 for correct answer, 0 otherwise. Answer can be simplified, but this is
not required
(a) Differentiate y = ex (x3 − 1).
Answer: y 0 = ex (x3 − 1) + 3x2 ex
Solution: Use the Product Rule.
(b) Differentiate C(q) =
1+q
with respect to q.
1 + q2
Answer:
(1 + q 2 ) − 2q(1 + q)
(1 + q 2 )2
Solution: Use the quotient rule.
(1 + q 2 ) − 2q(1 + q)
1 − 2q − q 2
0
C (q) =
=
(1 + q 2 )2
(1 + q 2 )2
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Determine the value of b for which f (x) =
x2 − 1, x < 3
is continuous at x = 3.
b + x, x ≥ 3
Answer: b = 5
Solution: Marking scheme: 1pt for correct answer. 1pt for some mention that
f (3) must equal to lim f (x)
x→3
We need
lim f (x) = f (3) = lim+ f (x)
x→3−
x→3
2
lim x − 1 = b + 3 = lim b + x
x→3
x→3
8=b+3
From here we see that b = 5.
(b) Find the equation of the tanget line to y =
1
at x = 1.
x2
Answer: y = 1 − 2(x − 1)
Solution: To find the slope at the point (1, 1), we need to evaluate the derivative
y 0 = −2x−3 at x = 1 to get m = −2. Therefore, the equation of the tangent line is
y = 1 − 2(x − 1) = −2x + 3.
Long answer question — you must show your work
3. 4 marks Use the definition
of the derivative, i.e. the limit process, to find the slope of the
√
tangent line to y = 2x + 3 at x = 3. Please put a box around your final answer.
Solution: Marking scheme: 1pt if correct answer but no limits. 1pt for limit definition;
1pt for rationalizing the numerator properly. 3pt for near perfect solution with small
algebra mistake.
f (x) − f (3)
x−3
√
√
2x + 3 − 6 + 3
= lim
x→3
x−3
√
√
2x + 3 − 3
2x + 3 + 3
·√
= lim
x→3
x−3
2x + 3 + 3
(2x + 3) − 9
√
= lim
x→3 (x − 3) · ( 2x + 3 + 3)
2x − 6
√
= lim
x→3 (x − 3) · ( 2x + 3 + 3)
2
(x−3)
√
= lim
x→3 (x−3)
· ( 2x + 3 + 3)
2
= lim √
x→3
2x + 3 + 3
2
2
1
=√
= =
6
3
9+3
y 0 = lim
x→3
f (3 + h) − f (3)
h→0
h
p
√
2(3 + h) + 3 − 6 + 3
= lim
h→0
h
√
9 + 2h − 3
= lim
h→0
h
√
√
9 + 2h − 3
9+h+3
= lim
·√
h→0
h
9+h+3
(9 + 2h) − 9
√
= lim
h→0 h · ( 9 + h + 3)
2
h
√
= lim
h→0 h · ( 9 + h + 3)
2
= lim √
h→0
9+h+3
2
2
1
=√
= =
6
3
9+3
= lim
104/184 Quiz #2
October 8
Grade:
First Name:
Last Name:
Student-No:
Section:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes.
scheme: 1 for each correct, 0 otherwise
Marking
(a) Differentiate f (t) = t−3 + et with respect to t.
Answer: f 0 (t) = −3t−4 + et
Solution: Use power rule (for negative integers) and exponential derivative.
(b) Let g(x) = x3 + 1. For what value(s) of x is the slope of the tangent line of g(x) equal to
12?
Answer: x = +2 and x = −2
Solution: Differentiate g(x) to get g 0 (x) = 3x2 . Setting 3x2 = 12 gives x2 = 4, so
x = ±2.
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Determine the value of c for which f (x) =
1/x, x ∈ (0, 2]
is continuous at x = 2.
cx,
x>2
Answer: c =
1
4
Marking scheme: 1pt for correct answer. 1pt for some mention that f (2) must equal
to lim f (x)
x→2
Solution: We need
lim f (x) = f (2) = lim+ f (x)
x→2−
1
1
=
x→2 x
2
1
1
=
2
2
lim
x→2
= lim cx
x→2
= 2c
Hence c = 14 .
(b) Differentiate y =
xex
. You do not need to simplify the answer.
x−2
Answer:
(ex +xex )(x−2)−(xex )
(x−2)2
Solution: Marking scheme: 1pt for correct use of product rule, 1pt for correct use
of quotient rule.
(xex )0 (x − 2) − (x − 2)0 (xex )
(x − 2)2
(ex + xex )(x − 2) − (1)(xex )
=
(x − 2)2
x 2
e (x − 2x − 2)
=
(x − 2)2
y0 =
Long answer question — you must show your work
3. 4 marks Use the definition of the derivative, i.e. the limit process, to compute the derivative
of
1
f (x) =
2x + 1
at x = 3. Please put a box around your final answer.
Solution: Marking scheme: 1pt if correct answer but no limits. 1pt for limit definition;
1pt for common-denominatoring. 3pt for near perfect solution with small algebra mistake.
f (x) − f (3)
x→3
x−3
1
− 71
2x+1
= lim
x→3 x − 3
1
7 − (2x + 1)
= lim
·
x→3
x−3
7 · (2x + 1)
6 − 2x
1
= lim
·
x→3
x−3
7 · (2x + 1)
1
−2
(x−3)
= lim
·
x→3 (x−3)
7 · (2x + 1)
f 0 (3) = lim
=
−2
72
f (3 + h) − f (3)
h→0
h
1
− 17
2(3+h)+1
= lim
h→0
h
1
7 − (2(3 + h) + 1)
= lim
·
h→0
h
7 · (2(3 + h) + 1)
1
7 − 6 − 2h − 1
= lim
·
h→0
h
7 · (2(3 + h) + 1)
1
−2
h
= lim
·
h→0 h
7 · (2(3 + h) + 1)
= lim
=
−2
72
104/184 Quiz #2
October 8
Grade:
First Name:
Last Name:
Student-No:
Section:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes.
Marking scheme: 1 for correct answer, 0 otherwise. Answer can be simplified, but this is
not required.
1
(a) Find the equation of the tangent line to y = x3 at x = .
2
Answer: y = 18 + 43 x − 12
Solution: The point is 21 , 18 . To find the slope, we need to evaluate the derivative
y 0 = 3x2 at x = 1/2 to get m = 34 .
Therefore, the equation of the tangent line is y = 18 + 43 x − 12 .
(b) Differentiate C(q) = (q − 1)(eq + 1) with respect to q.
Answer: C 0 = (eq + 1) + (q − 1)eq
Solution: Use the product rule: C 0 (q) = 1(eq + 1) + (q − 1)eq = qeq + 1
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Determine the value of a for which f (x) =
ex ,
x<2
is continuous at x = 2.
1 + ax2 , x ≥ 2
e2 − 1
Answer: a =
4
Solution: Marking scheme: 1pt for correct answer. 1pt for some mention that
f (2) must equal to lim f (x)
x→2
We need
lim f (x) = f (2)
x→2−
= lim+ f (x)
x→2
x
lim e = 1 + 4a = lim 1 + ax2
x→2
x→2
2
e = 1 + 4a = 1 + 4a
From here we see that e2 = 1 + 4a ⇒ a =
e2 − 1
.
4
(b) Differentiate y =
ex + x
. You do not need to simplify the answer.
x3 + 1
Answer:
(ex +1)(x3 +1)−(3x2 )(ex +x)
(x3 +1)2
Solution: Marking scheme: 2pt for correct answer. 1pt for some work in the right
direction
(ex + x)0 (x3 + 1) − (x3 + 1)0 (ex + x)
(x3 + 1)2
(ex + 1)(x3 + 1) − (3x2 )(ex + x)
=
(x3 + 1)2
ex (x3 − 3x2 + 1) + (−2x3 + 1)
=
(x3 + 1)2
y0 =
Long answer question — you must show your work
3. 4 marks Use the definition of the derivative, i.e. the limit process, to find the slope of the
tangent line to
√
y = 5x − 1
at x = 2. Please put a box around your final answer.
Solution: Marking scheme: 1pt if correct answer but no limits. 1pt for limit definition;
1pt for rationalizing the numerator properly. 3pt for near perfect solution with small
algebra mistake.
f (x) − f (2)
x→2
x−2
√
√
5x − 1 − 10 − 1
= lim
x→2
x−2
√
√
5x − 1 − 3
5x − 1 + 3
= lim
·√
x→2
x−2
5x − 1 + 3
(5x − 1) − 9
√
= lim
x→2 (x − 2) · ( 5x − 1 + 3)
5x − 10
√
= lim
x→2 (x − 2) · ( 5x − 1 + 3)
5
(x−2)
= lim
√
x→2 (x−2)
· ( 5x − 1 + 3)
5
= lim √
x→2
5x − 1 + 3
5
5
=
=
3+3
6
y 0 = lim
f (2 + h) − f (2)
h→0
h
p
√
5(2 + h) − 1 − 10 − 1
= lim
h→0
h
√
9 + 5h − 3
= lim
h→0
h
√
√
9 + 5h − 3
9 + 5h + 3
= lim
·√
h→0
h
9 + 5h + 3
(9 + 5h) − 9
√
= lim
h→0 h · ( 9 + 5h + 3)
5
h
√
= lim
h→0 h · ( 9 + 5h + 3)
5
= lim √
h→0
9 + 5h + 3
5
5
=
=
3+3
6
= lim
104/184 Quiz #2
October 8
First Name:
Last Name:
Student-No:
Section:
Grade:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes. You do not
need to simplify the answer. Marking scheme: 1 for each correct, 0 otherwise
x3
(a) Differentiate y = x .
e
Answer:
(3x2 )(ex )−(ex )(x3 )
(ex )2
Solution:
(x3 )0 (ex + 1) − (ex + 1)0 (x3 )
(ex + 1)2
(3x2 )(ex + 1) − (ex )(x3 )
=
(ex + 1)2
ex (3x2 − x3 ) + 3x2
=
(ex + 1)2
y0 =
(b) Differentiate C(q) = q 3 eq with respect to q.
Answer: C 0 (q) = 3q 2 eq + q 3 eq
Solution: Use the product rule.
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
√
a x,
x<2
(a) Determine the value of a for which f (x) =
is continuous at x = 2.
2
1+x , x≥2
Answer: a =
√5
2
Solution: Marking scheme: 1pt for correct answer. 1pt for some mention that
f (2) must equal to lim f (x)
x→2
We need
lim f (x) = f (2) = lim+ f (x)
x→2
√
2
lim a x = 1 + 2 = lim 1 + x2
x→2
x→2
√
a 2=5
= 1 + 22
x→2−
√
5
From here we see that a 2 = 5 ⇒ a = √ .
2
(b) Find the equation of the tangent line to y = x3 at x = 13 .
1
1
Answer: y =
+
27 3
1
x−
3
Solution: Marking scheme: 2pt for correct answer. 1pt for some work in the right
direction
To find the slope at the point (1, 1/27), we need to evaluate the derivative y 0 = 3x2
at x = 1/3 to
m = 1/3. Therefore, the equation of the tangent line is y =
get
1
1
1
2
1
+
x
−
=
x
− 27
.
27
3
3
3
Long answer question — you must show your work
3. 4 marks Use the definition of the derivative, i.e. the limit process, to find the slope of the
tangent line to
1
y=
2x + 3
at x = 2. Please put a box around your final answer. Marking scheme: 1pt if correct
answer but no limits. 1pt for limit definition; 1pt for correct rearrangement of fractions. 3pt
for near perfect solution with small algebra mistake.
Solution:
f (x) − f (2)
x→2
x−2
1
1
− 2(2)+3
2x+3
= lim
x→2
x−2
1
− 17
2x+3
= lim
x→2 x − 2
y 0 = lim
= lim
7−(2x+3)
7(2x+3)
x−2
4 − 2x
= lim
x→2 7(2x + 3)(x − 2)
−2
(x−2)
= lim
x→2 7(2x + 3)
(x−2)
x→2
=
−2
−2
=
7(4 + 3)
49