104/184 Quiz 2
October 9
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
(a) Differentiate C(q) = q 3 eq with respect to q..
Answer: 3q 2 eq + q 3 eq
Solution: C 0 (q) = (q 3 )0 eq + q 3 (eq )0 = 3q 2 eq + q 3 eq
(b) Differentiate y =
x5
.
ex + 3
Answer:
(5x4 )(ex + 3) − x5 (ex )
(ex + 3)2
Solution:
(x5 )0 (ex + 3) − x5 (ex + 3)0
(ex + 3)2
4
x
(5x )(e + 3) − x5 (ex )
=
(ex + 3)2
y0 =
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(
1 + eax , x < 4
(a) Determine the value of a for which f (x) = √
is continuous at x = 4.
x,
x≥4
Answer: a = 0
Solution: Marking scheme: 1pt for correct answer. 1pt for some mention that
f (4) must equal to lim f (x)
x→4
lim f (x) = f (4) = lim+ f (x)
x→4
√
√
ax
lim− (1 + e ) = 4 = lim+ x
x→4
x→4
√
4a
1+e =2= 4
x→4−
From here we see that e4a = 1 ⇒ a = 0.
(b) Find the equation of the tangent line to y = 2x4 at x = −1.
Answer: y = 2 − 8(x + 1)
Solution: Marking scheme: 1pt for correct answer; 1pt for some work in the right
direction
To find the slope at the point (−1, 2), we need to evaluate the derivative y 0 = 8x3
at x = −1 to get m = −8. Therefore, the equation of the tangent line is y =
−8(x − (−1)) + 2 = −8x − 6.
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
tagent line to the graph of
2
y=
3x + 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 using the common denominator. 3pt for near perfect solution with small algebra
mistake.
2
2
− 3(2)+1
f (x) − f (2)
3x+1
= lim
f (2) = lim
x→2
x→2
x−2 x−2
2
2
− [7(3x + 1)]
2(7) − 2(3x + 1)
= lim 3x+1 7
= lim
x→2
x→2 7(x − 2)(3x + 1)
(x − 2)[7(3x + 1)]
−6(x − 2)
2(6 − 3x)
= lim
= lim
x→2 7(x − 2)(3x + 1)
x→2 7(x − 2)(3x + 1)
−6
6
−6
=
=−
lim
x→2 7(3x + 1)
7(3(2) + 1)
49
0
or
2
−
f (2 + h) − f (2)
3(2+h)+1
f (2) = lim
= lim
h→0
h→0
h h
2
2
−
[7(3h
+
7)]
= lim 3h+7 7
h→0
h[7(3h + 7)]
2(7) − 2(3h + 7)
= lim
h→0
7h(3h + 7)
−6h
= lim
h→0 7h(3h + 7)
−6
−6
6
= lim
=
=−
h→0 7(3h + 7)
7(3(0) + 7)
49
0
2
3(2)+1
104/184 Quiz #2
October 9
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. You do not
need to simplify the answer. Marking scheme: 1 for each correct, 0 otherwise
ex
.
(a) Differentiate y =
2x + 1
ex (2x + 1) − 2ex
Answer:
(2x + 1)2
Solution:
(ex )0 · (2x + 1) − (2x + 1)0 · (ex )
(2x + 1)2
e2x (2x + 1) − 2ex
=
(2x + 1)2
y0 =
(b) Differentiate C(q) = q −2 eq with respect to q.
Answer: C 0 (q) = −2q −3 eq + q −2 eq
Solution: Use the product rule.
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) =
ln(ax) + 2, x < 2
is continuous at x = 2.
3x − 4,
x≥2
Answer: a =
1
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)
x→2−
=
f (2)
=
lim f (x)
x→2+
lim (ln(ax) + 2) = 3 × 2 − 4 = lim (3x − 4)
x→2−
ln(2a) + 2
x→2
=
2
=
From here we see that ln(2a) = 0 ⇒ 2a = e0 ⇒ a =
1
.
2
3×2−4
(b) Find the equation of the tanget line to y = 3x2 + x at x = 2.
Answer: y = 14 + 13(x − 2)
Solution: Marking scheme: 1pt for correct answer; 1pt for some work in the right
direction
The point is (2, 14). To find the slope, we need to evaluate the derivative y 0 = 6x + 1
at x = 2 to get m = 13.
Therefore, the equation of the tanget line is y = 14 + 13(x − 2) = 13x − 12.
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. Marking scheme: 1pt if correct
answer but no limits. 1pt for limit definition; 1pt for rationalizing. 3pt for near perfect solution
with small algebra mistake.
Solution:
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
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
h→0
104/184 Quiz #2
October 9
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. You do not
need to simplify the answer.
(a) Differentiate P (q) = (q + 1)eq with respect to q.
Answer: P 0 (q) = eq + (q + 1)eq
Marking scheme: 1 for correct answer, 0 otherwise
Solution: Use the product rule P 0 (q) = eq + (q + 1)eq = (q + 2)eq .
(b) Differentiate y =
1 + x2
.
1 + ex
Answer:
(1 + ex )(2x) − (1 + x2 )(ex )
(1 + ex )2
Solution: Use the quotient rule:
y0 =
(1 + ex )(2x) − (1 + x2 )(ex )
(1 + ex )2
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) =
1 + x + x2 , x < 1
is continuous at x = 1.
e−ax ,
x≥1
Answer: a = − ln(3)
Solution: Marking scheme: 1pt for correct answer. 1pt for some mention that
f (1) must equal to lim f (x)
x→1
Compute
lim f (x) = lim+ e−ax = e−a
x→1+
x→1
and
lim f (x) = lim− 1 + x + x2 = 3
x→1−
x→1
and f (1) = e−a . All these values must be equal therefore e−a = 3 and so a = − ln(3).
(b) Find the equation of the tangent line to y = x13 at x = −1.
Answer: y = −1 + 13(x + 1)
Solution: Marking scheme: 1pt for correct answer; 1pt for some work in the right
direction
To find the slope at the point (−1, −1), we need to evaluate the derivative y 0 =
13x12 at x = −1 to get m = 13. Therefore, the equation of the tangent line is
y = 13(x − (−1)) − 1 = 13x + 12.
Long answer question — you must show your work
3. 4 marks Use the definition of the derivative, ie. 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.
Solution: Marking scheme: 1pt if correct answer but no limits. 1pt for limit definition;
1pt for using the common denominator. 3pt for near perfect solution with small algebra
mistake.
f (x) − f (2)
x→2
x−2
1
−1
= lim 2x−3
x→2 x − 2
f 0 (2) = lim
= lim
1−(2x−3)
2x−3
x−2
4 − 2x
= lim
x→2 (2x − 3)(x − 2)
−2
(x−2)
= lim
x→2 (2x − 3)
(x−2)
−2
=
1
= −2
x→2
f (2 + h) − f (2)
h→0
h
1
−1
2(2+h)−3
= lim
h→0
h
= lim
= lim
1−(4+2h−3)
2(2+h)−3
h
1 − 1 − 2h
= lim
h→0 h · (2(2 + h) − 3)
−2
h
= lim
h→0 h · (2h + 1)
−2
=
1
= −2
h→0
104/184 Quiz #2
October 9
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
(a) 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.
(b) Differentiate y =
ex
x
.
+ x2
Answer:
(ex + x2 ) − x(ex + 2x)
(ex + x2 )2
Solution:
(x)0 (ex + x2 ) − x(ex + x2 )0
(ex + x2 )2
(ex + x2 ) − x(ex + 2x)
=
.
(ex + x2 )2
y0 =
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Find the equation of the tanget line to y = x3 at x = −1.
Answer: y = −1 + 3(x + 1)
Solution: Marking scheme: 1pt for correct answer; 1pt for some work in the right
direction
The point is (−1, −1). To find the slope, we need to evaluate the derivative y 0 = 3x2
at x = −1 to get m = 3.
Therefore, the equation of the tanget line is y = −1 + 3(x + 1) = 3x + 2.
√
x + 8, x < 1
(b) Determine the value of a for which f (x) =
is continuous at x = 1.
eax ,
x≥1
Answer: a = ln 3
Solution: Marking scheme: 1pt for correct answer. 1pt for some mention that
f (1) must equal to lim f (x)
x→1
We need
lim f (x) = f (1) = lim+ f (x)
x→1
√
a
= lim eax
lim x + 8 = e
x→1−
x→1
a
3=e
x→1
a
=e
¿From here we see that ea = 3 ⇒ a = ln 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 = 4x + 1
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
rationalizing. 3pt for near perfect solution with small algebra mistake.
Solution:
f (x) − f (2)
x→2
x−2
√
√
4x + 1 − 8 + 1
= lim
x→2
x−2
√
√
4x + 1 − 3
4x + 1 + 3
= lim
·√
x→2
x−2
4x + 1 + 3
(4x + 1) − 9
√
= lim
x→2 (x − 2) · ( 4x + 1 + 3)
4x − 8
√
= lim
x→2 (x − 2) · ( 4x + 1 + 3)
4
(x−2)
= lim
√
x→2 (x−2)
· ( 4x + 1 + 3)
4
= lim √
x→2
4x + 1 + 3
4
4
2
=√
= =
6
3
9+3
y 0 = lim
f (2 + h) − f (2)
h→0
h
p
√
4(2 + h) + 1 − 8 + 1
= lim
h→0
h
√
9 + 4h − 3
= lim
h→0
h
√
√
9 + 4h − 3
9 + 4h + 3
= lim
·√
h→0
h
9 + 4h + 3
(9 + 4h) − 9
√
= lim
h→0 h · ( 9 + 4h + 3)
4
h
√
= lim
h→0 h · ( 9 + 4h + 3)
4
= lim √
h→0
9 + 4h + 3
4
4
2
=√
= =
6
3
9+3
= lim