104/184 Quiz #4
November 20
First Name:
Last Name:
Student-No:
Section:
Grade:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Below is a graph of f 0 (x), the derivative of f (x).
Marking scheme: 1 for each correct, 0 otherwise. Accuracy is not required, but should be
within the correct 0.5 subinterval.
(a) When is f (x) increasing?
Answer: (−1, 2)
S
(3, ∞)
Solution: f (x) is increasing ⇔ f 0 (x) > 0.
(b) When does f (x) have an inflection point?
Answer: x = 0.2 or x = 2.5
Solution: When f 0 (x) changes from ‘increasing to decreasing or vise-versa; i.e. at
about x = 0.2 and x = 2.5.
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) When is the function g(x) = 3x4 + 4x3 concave up?
Answer:
(−∞, −2/3) ∪ (0, ∞)
Solution: Marking scheme: 1pt for a correct first line of the second derivative.
1pt for a correct final answer
We’ll need the second derivative:
g(x) = 3x4 + 4x3
g 0 (x) = 12x3 + 12x2
g 00 (x) = 36x2 + 24x
g 00 (x) = 12x(3x + 2)
Therefore g 00 (x) is a parabola opening upward with roots x = −2/3 or x = 0. Hence
g 00 (x) > 0 if and only if x < −2/3 or x > 0.
x2 − 9
. Determine whether
(2x − 7)
there is a local maximum or minimum value (or neither) at x = 3.
Answer: local maximum
(b) Suppose that f (x) is a function with derivative f 0 (x) =
Solution: Marking scheme: 1pt for the correct answer. 1pt for a valid justification.
Using the first derivative test. We see that the numerator changes from ‘-’ to ‘+’ at
x = 3 and the denominator is negative. Hence the first derivative changes from ‘+’
to ‘-’ and there must be a local maximum by the First Derivative Test.
Alternatively, we may use the Second Derivative Test. But first, we’ll need the second
derivative:
x2 − 9
(2x − 7)
2x(2x − 7) − 2(x2 − 9)
f 00 (x) =
(2x − 7)2
4x2 − 14x − 2x2 + 18
=
(2x − 7)2
2
2x − 14x + 18
=
(2x − 7)2
f 0 (x) =
18 − 42 + 18
= −6 < 0. From which we can see that f (3) is a
(6 − 7)2
local maximum value by the Second Derivative Test.
Therefore, f 00 (3) =
Long answer question — you must show your work
3. 4 marks Consider the function
√
f (x) = (x − 3) x,
which has derivatives
3 x−1
3 x+1
· √ and f 00 (x) = · 3/2 .
2
4 x
x
(a) When is f(x) increasing and decreasing?
f 0 (x) =
Solution: Marking scheme: 1pt for identifying when f 0 (x) > 0 and f 0 (x) < 0.
Justification is not required.
The denominator of f 0 (x) is alway positive (on its domain), so the sign of f 0 (x) is
completely determined by x − 1. Therefore,
f (x) is increasing ⇔ f 0 (x) > 0 ⇔ x > 1,
and
f (x) is decreasing ⇔ f 0 (x) < 0 ⇔ x < 1.
(b) When is f(x) concave up or concave down?
Solution: Marking scheme: 1pt for identifying when f 00 (x) > 0 and f 00 (x) < 0.
Justification is not required. Note that the domain of f (x) is [0, ∞). Thus, f 00 (x) > 0
whenever x > 0. Hence, it’s always concave up!
(c) Sketch the graph of f (x). Clearly label any extreme values and inflection points.
Solution: Marking scheme: 2pts. 1pt clearly labeling the minimum and 1pt sketcing the graph on the proper domain.
The intercepts are given by:
x-intercepts: (0, 0) and (3, 0). y-intercept: (0, 0)
The only critical numbers are x = 0 and x = 1, and there are NO possible inflection
points. It will be helpful to tabulate all the information. We’ll break up the domain
at each important point.
Function\Interval
f (x)
f 0 (x)
f 00 (x)
(0, 1) (1, 3) (3, ∞)
+
+
+
+
+
+
Consequently, there must be a local min at (1, −2), with no local max or inflection
points.
The graph should look something like this:
104/184 Quiz #4
November 20
First Name:
Last Name:
Student-No:
Section:
Grade:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Below is a graph of f 0 (x), the derivative of f (x).
Marking scheme: 1 for each correct, 0 otherwise
(a) Where is f (x) decreasing?
Answer: (−∞, −1)
S
(2, 3)
Solution: f (x) is decreasing ⇔ f 0 (x) < 0.
(b) Where does f (x) have an inflection point?
Answer: x = 0.2 or x = 2.5
Solution: When f 0 (x) changes from ‘increasing to decreasing or vise-versa; i.e. at
about x = o.2 and x = 2.5.
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) When is the function g(x) = 3x4 + 4x3 concave down?
Answer: (−2/3, 0)
Solution: Marking scheme: 1pt for a correct first line of differentiation. 1pt for a
correct final answer We’ll need the second derivative:
g(x) = 3x4 + 4x3
g 0 (x) = 12x3 + 12x2
g 00 (x) = 36x2 + 24x
g 00 (x) = 12x(3x + 2)
Therefore g 00 (x) is a parabola opening upward with roots x = −2/3 or x = 0. Hence
g 00 (x) < 0 if and only if −2/3 < x < 0.
x2 − 9
. Determine whether
(2x − 7)
there is a local maximum or minimum value (or neither) at x = −3.
Answer: local minimum
(b) Suppose that f (x) is a function with derivative f 0 (x) =
Solution: Marking scheme: 1pt for the correct answer. 1pt for a valid justification.
Using the first derivative test. We see that the numerator changes from ‘+’ to ‘-’ at
x = −3 and the denominator is negative. Hence the first derivative changes from ‘-’
to ‘+’ and there must be a local minimum by the First Derivative Test.
Alternatively, we may use the Second Derivative Test. But first, we’ll need the second
derivative:
x2 − 9
(2x − 7)
2x(2x − 7) − 2(x2 − 9)
f 00 (x) =
(2x − 7)2
4x2 − 14x − 2x2 + 18
=
(2x − 7)2
2x2 − 14x + 18
=
(2x − 7)2
f 0 (x) =
18 + 42 + 18
=> 0. From which we can see that f (−3) is a local
(−6 − 7)2
minimum value by the Second Derivative Test.
Therefore, f 00 (−3) =
Long answer question — you must show your work
3. 4 marks Consider the function
√
f (x) = (x − 6) x,
which has derivatives
3 x−2
3 x+2
· √ and f 00 (x) = · 3/2 .
2
4 x
x
(a) When is f(x) increasing and decreasing?
f 0 (x) =
Solution: Marking scheme: 1pt for identifying when f 0 (x) > 0 and f 0 (x) < 0.
Justification is not required.
The denominator of f 0 (x) is alway positive (on its domain), so the sign of f 0 (x) is
completely determined by x − 2. Therefore,
f (x) is increasing ⇔ f 0 (x) > 0 ⇔ x > 2,
and
f (x) is decreasing ⇔ f 0 (x) < 0 ⇔ x < 2.
(b) When is f(x) concave up or concave down?
Solution: Marking scheme: 1pt for identifying when f 00 (x) > 0 and f 00 (x) < 0.
Justification is not required. Note that the domain of f (x) is [0, ∞). Thus, f 00 (x) > 0
whenever x > 0. Hence, it’s always concave up!
(c) Sketch the graph of f (x). Clearly label any extreme values and inflection points.
Solution: Marking scheme: 2pts. 1pt clearly labeling the minimum and 1pt sketcing the graph on the proper domain.
The intercepts are given by:
x-intercepts: (0, 0) and (6, 0). y-intercept: (0, 0)
The only critical numbers are x = 0 and x = 2, there are NO possible inflection
points. It will be helpful to tabulate all the information. We’ll break up the domain
at each important point.
(0, 2) (2, 6) (6, ∞)
+
+
+
+
+
+
√
Consequently, there must be a local min at (2, −4 2), with no local max or inflection
points.
The graph should look something like this:
Function\Interval
f (x)
f 0 (x)
f 00 (x)
104/184 Quiz #4
November 20
First Name:
Last Name:
Student-No:
Section:
Grade:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Below is a graph of f 0 (x), the derivative of f (x).
Marking scheme: 1 for each correct, 0 otherwise
(a) When is f (x) increasing?
Answer: (−∞, −1)
S
(2, 3)
Solution: f (x) is increasing ⇔ f 0 (x) > 0.
(b) When does f (x) have an inflection point?
Answer: x = 0.2 or x = 2.5
Solution: When f 0 (x) changes from ‘increasing to decreasing or vise-versa; i.e. at
about x = 0.2 and x = 2.5.
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) When is the function g(x) = 3x4 − 4x3 concave up?
Answer:
(−∞, 0) ∪ (2/3, ∞)
Solution: Marking scheme: 1pt for a correct first line of the second derivative.
1pt for a correct final answer
We’ll need the second derivative:
g(x) = 3x4 − 4x3
g 0 (x) = 12x3 − 12x2
g 00 (x) = 36x2 − 24x
g 00 (x) = 12x(3x − 2)
Therefore g 00 (x) is a parabola opening upward with roots x = 0 or x = 2/3. Hence
g 00 (x) > 0 if and only if x < 0 or x > 2/3.
9 − x2
. Determine whether
(2x − 7)
there is a local maximum or minimum value (or neither) at x = 3.
(b) Suppose that f (x) is a function with derivative f 0 (x) =
Answer: local minimum
Solution: Marking scheme: 1pt for the correct answer. 1pt for a valid justification.
Using the first derivative test. We see that the numerator changes from ‘+’ to ‘-’ at
x = 3 and the denominator is negative. Hence the first derivative changes from ‘-’ to
‘+’ and there must be a local minimum by the First Derivative Test.
Alternatively, we may use the Second Derivative Test. But first, we’ll need the second
derivative:
9 − x2
(2x − 7)
−2x(2x − 7) − 2(9 − x2 )
f 00 (x) =
(2x − 7)2
2
−4x + 14x − 18 + 2x2
=
(2x − 7)2
−2x2 + 14x − 18
=
(2x − 7)2
f 0 (x) =
−18 + 42 − 18
= 6 > 0. From which we can see that f (3) is a
(6 − 7)2
local minimum value by the Second Derivative Test.
Therefore, f 00 (3) =
Long answer question — you must show your work
3. 4 marks Consider the function
√
f (x) = x x + 3,
which has derivatives
f 0 (x) =
3 x+2
3
x+4
·√
and f 00 (x) = ·
.
2
4 (x + 3)3/2
x+3
(a) When is f(x) increasing and decreasing?
Solution: Marking scheme: 1pt for identifying when f 0 (x) > 0 and f 0 (x) < 0.
Justification is not required.
The denominator of f 0 (x) is alway positive (on its domain), so the sign of f 0 (x) is
completely determined by x + 2. Therefore,
f (x) is increasing ⇔ f 0 (x) > 0 ⇔ x > −2,
and
f (x) is decreasing ⇔ f 0 (x) < 0 ⇔ x < −2.
(b) When is f(x) concave up or concave down?
Solution: Marking scheme: 1pt for identifying when f 00 (x) > 0 and f 00 (x) < 0.
Justification is not required.
Note that the domain of f (x) is [−3, ∞). Thus,
f 00 (x) > 0 whenever x > −3. Hence, it’s always concave up!
(c) Sketch the graph of f (x). Clearly label any extreme values and inflection points.
Solution: Marking scheme: 2pts. 1pt clearly labeling the minimum and 1pt sketcing the graph on the proper domain.
The intercepts are given by:
x-intercepts: (−3, 0) and (0, 0). y-intercept: (0, 0)
The only critical points are x = −3 and x = −2, and there are NO possible inflection
points. It will be helpful to tabulate all the information. We’ll break up the domain
at each important point.
Function\Interval
f (x)
f 0 (x)
f 00 (x)
(−3, −2) (−2, 0) (0, ∞)
+
+
+
+
+
+
Consequently, there must be a local min at (−2, −2), with no local max or inflection
points.
The graph should look something like this:
104/184 Quiz #4
November 20
First Name:
Last Name:
Student-No:
Section:
Grade:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Below is a graph of f 0 (x), the derivative of f (x).
Marking scheme: 1 for each correct, 0 otherwise. Accuracy is not required, but should be
within the correct 0.5 subinterval.
(a) When is f (x) decreasing?
Answer: (−1, 2)
S
(3, ∞)
Solution: f (x) is decreasing ⇔ f 0 (x) < 0.
(b) When does f (x) have an inflection point?
Answer: x = 0.2 or x = 2.5
Solution: When f 0 (x) changes from ‘increasing to decreasing or vise-versa; i.e. at
about x = 0.2 and x = 2.5.
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) When is the function g(x) = 3x4 − 4x3 concave down?
Answer: (0, 2/3)
Solution: Marking scheme: 1pt for a correct first line of the second derivative.
1pt for a correct final answer
We’ll need the second derivative:
g(x) = 3x4 − 4x3
g 0 (x) = 12x3 − 12x2
g 00 (x) = 36x2 − 24x
g 00 (x) = 12x(3x − 2)
Therefore g 00 (x) is a parabola opening upward with roots x = 0 or x = 2/3. Hence
g 00 (x) < 0 if and only if 0 < x < 2/3.
9 − x2
. Determine whether
(2x − 7)
there is a local maximum or minimum value (or neither) at x = −3.
(b) Suppose that f (x) is a function with derivative f 0 (x) =
Answer: local maximum
Solution: Marking scheme: 1pt for the correct answer. 1pt for a valid justification.
Using the first derivative test. We see that the numerator changes from ‘-’ to ‘+’ at
x = −3 and the denominator is negative. Hence the first derivative changes from ‘+’
to ‘-’ and there must be a local maximum by the First Derivative Test.
Alternatively, we may use the Second Derivative Test. But first, we’ll need the second
derivative:
9 − x2
(2x − 7)
−2x(2x − 7) − 2(9 − x2 )
f 00 (x) =
(2x − 7)2
2
−4x + 14x − 18 + 2x2
=
(2x − 7)2
−2x2 + 14x − 18
=
(2x − 7)2
f 0 (x) =
−18 − 42 − 18
< 0. From which we can see that f (−3) is a
(6 − 7)2
local maximum value by the Second Derivative Test.
Therefore, f 00 (−3) =
Long answer question — you must show your work
3. 4 marks Consider the function
√
f (x) = (x) x + 6,
which has derivatives
f 0 (x) =
3 x+4
3
x+8
·√
and f 00 (x) = ·
.
2
4 (x + 6)3/2
x+6
(a) When is f(x) increasing and decreasing?
Solution: Marking scheme: 1pt for identifying when f 0 (x) > 0 and f 0 (x) < 0.
Justification is not required.
The denominator of f 0 (x) is alway positive (on its domain), so the sign of f 0 (x) is
completely determined by x + 4. Therefore,
f (x) is increasing ⇔ f 0 (x) > 0 ⇔ x > −4,
and
f (x) is decreasing ⇔ f 0 (x) < 0 ⇔ x < −4.
(b) When is f(x) concave up or concave down?
Solution: Marking scheme: 1pt for identifying when f 00 (x) > 0 and f 00 (x) < 0.
Justification is not required.
Note that the domain of f (x) is [−6, ∞). Thus,
00
f (x) > 0 whenever x > −6. Hence, it’s always concave up!
(c) Sketch the graph of f (x). Clearly label any extreme values and inflection points.
Solution: Marking scheme: 2pts. 1pt clearly labeling the minimum and 1pt sketcing the graph on the proper domain.
The intercepts are given by:
x-intercepts: (−6, 0) and (0, 0). y-intercept: (0, 0)
The only critical points are x = −6 and x = −4, and there are NO possible inflection
points. It will be helpful to tabulate all the information. We’ll break up the domain
at each important point.
(−6, −4) (−4, 0) (0, ∞)
+
+
+
+
+
+
√
Consequently, there must be a local min at (−4, −4 2), with no local max or inflection
points.
The graph should look something like this:
Function\Interval
f (x)
f 0 (x)
f 00 (x)