104/184 Quiz #4
November 19
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) Where is f (x) concave up?
Answer: S
(−∞, 0.2)
(2.5, ∞)
Solution: f (x) concave up ⇔ f 00 (x) > 0 ⇔ f 0 (x) is increasing.
(b) Where does f (x) have a local maximum?
Answer: x = 2.
Solution: When f 0 (x) changes from ‘+’ to ‘-’.
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) When is the function g(x) = xe3x increasing?
Answer: x > −1/3
Solution: Marking scheme: 1pt for a correct first line of differentiation. 1pt for a
correct final answer We’ll need the derivative:
g(x) = xe3x
g 0 (x) = e3x + xe3x · 3
g 0 (x) = e3x (1 + 3x)
Therefore g 0 (x) > 0 if and only if x > −1/3.
(b) Suppose that f (x) is a function with f 0 (π) = 0 and f 00 (x) =
x2 − 9
. Determine
2x · (2x − 3)
whether there is a local maximum or minimum value at x = π.
Answer: local minimum
Solution: Marking scheme: 1pt for determining that f 00 (π) > 0. 1pt for a correct
final answer The value f 00 (π) is given by
f 00 (π) =
π2 − 9
>0
2π · (2π − 3)
since π > 3 implies that π 2 − 9 > 0. Therefore, f (π) is a local minimum value by the
Second Derivative Test.
Long answer question — you must show your work
3. 4 marks Consider the function
f (x) = (x + 1)(x − 2)2 ,
which has derivatives
f 0 (x) = 3x(x − 2) and f 00 (x) = 6(x − 1).
(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. f 0 (x) a parabola open upwards with roots x = 0 and
x = 2. Therefore,
f (x) is increasing ⇔ f 0 (x) > 0 ⇔ x < 0 or x > 2,
and
f (x) is decreasing ⇔ f 0 (x) < 0 ⇔ 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. f 00 (x) a line with a positve slope, we see that:
f (x) is concave up ⇔ f 00 (x) > 0 ⇔ x > 1,
and
f (x) is concave down ⇔ f 00 (x) < 0 ⇔ x < 1.
(c) Sketch the graph of f (x). Clearly label any extreme values and inflection points.
Solution: Marking scheme: 2pts. 1pt clearly labeling the extreme values and 1pt
for labelling the inflection point.
The intercepts are given by:
x-intercepts: (−1, 0) and (2, 0). y-intercept: (0, 2)
The critical numbers are x = 0 and x = 2, and the possible inflection point is at
x = 1. 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)
(−∞, −1) (−1, 0) (0, 1) (1, 2) (2, ∞)
+
+
+
+
+
+
+
+
+
Consequently, there must be a local max at (0, 4), local min at (2, 0), and an inflection
point when (1, 2).
The graph should look something like this:
104/184 Quiz #4
November 19
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) concave down?
Answer: (0.2, 2.5)
Solution: f (x) concave down ⇔ f 00 (x) < 0 ⇔ f 0 (x) is decreasing.
(b) Where does f (x) have a local minimum?
Answer: x = −1 or x = 3
Solution: When f 0 (x) changes from ‘-’ to ‘+’.
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) When is the function g(x) = xe4x increasing?
Answer: x > −1/4
Solution: Marking scheme: 1pt for a correct first line of differentiation. 1pt for a
correct final answer We’ll need the derivative:
g(x) = xe4x
g 0 (x) = e4x + xe4x · 4
g 0 (x) = e4x (1 + 4x)
Therefore g 0 (x) > 0 if and only if x > −1/4.
(b) Suppose that f (x) is a function with f 0 (π) = 0 and f 00 (x) =
9 − x2
. Determine
2x · (2x − 3)
whether there is a local maximum or minimum value at x = π.
Answer: local maximum
Solution: Marking scheme: 1pt for determining that f 00 (π) < 0. 1pt for a correct
final answer.
The value f 00 (π) is given by
f 00 (π) =
9 − π2
<0
2π · (2π − 3)
since π > 3 implies that 9 − π 2 < 0. Therefore, f (π) is a local maximum value by the
Second Derivative Test.
Long answer question — you must show your work
3. 4 marks Consider the function
f (x) = (x − 1)(x + 2)2 ,
which has derivatives
f 0 (x) = 3x(x + 2) and f 00 (x) = 6(x + 1).
(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. f 0 (x) a parabola open upwards with roots x = −2 and
x = 0. Therefore,
f (x) is increasing ⇔ f 0 (x) > 0 ⇔ x < −2 or x > 0,
and
f (x) is decreasing ⇔ f 0 (x) < 0 ⇔ −2 < x < 0.
(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. f 00 (x) a line with a positve slope, we see that:
f (x) is concave up ⇔ f 00 (x) > 0 ⇔ x > −1,
and
f (x) is concave down ⇔ f 00 (x) < 0 ⇔ x < −1.
(c) Sketch the graph of f (x). Clearly label any extreme values and inflection points.
Solution: Marking scheme: 2pts. 1pt clearly labeling the extreme values and 1pt
for labelling the inflection point.
The intercepts are given by:
x-intercepts: (−2, 0) and (1, 0). y-intercept: (0, −4)
The critical numbers are x = −2 and x = 0, and the possible inflection point is at
x = −1. 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)
(−∞, −2) (−2, −1) (−1, 0) (0, 1) (1, ∞)
+
+
+
+
+
+
+
Consequently, there must be a local max at (−2, 0), local min at (0, −4), and an
inflection point when (−1, −2).
The graph should look something like this:
104/184 Quiz #4
November 19
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) Where is f (x) concave up?
Answer: (−0.2, 2.5)
Solution: f (x) concave up ⇔ f 00 (x) > 0 ⇔ f 0 (x) is increasing.
(b) Where does f (x) have a local maximum?
Answer: x = −1 or x = 3
Solution: When f 0 (x) changes from ‘+’ to ‘-’.
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) When is the function g(x) = xe5x increasing?
Answer: x > −1/5
Solution: Marking scheme: 1pt for a correct first line of differentiation. 1pt for a
correct final answer We’ll need the derivative:
g(x) = xe5x
g 0 (x) = e5x + xe5x · 5
g 0 (x) = e5x (1 + 5x)
Therefore g 0 (x) > 0 if and only if x > −1/5.
(b) Suppose that f (x) is a function with f 0 (π) = 0 and f 00 (x) =
x2 + 9
. Determine
2x · (2x − 6)
whether there is a local maximum or minimum value at x = π.
Answer: local minimum
Solution: Marking scheme: 1pt for determining that f 00 (π) > 0. 1pt for a correct
final answer
The value f 00 (π) is given by
f 00 (π) =
π2 + 9
>0
2π · (2π − 6)
since π > 3 implies that 2π − 6 > 0. Therefore, f (π) is a local minimum value by the
Second Derivative Test.
Long answer question — you must show your work
3. 4 marks Consider the function
f (x) = (x + 1)2 (x − 2),
which has derivatives
f 0 (x) = 3(x2 − 1) and f 00 (x) = 6x.
(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. f 0 (x) a parabola open upwards with roots x = −1 and
x = 1. Therefore,
f (x) is increasing ⇔ f 0 (x) > 0 ⇔ x < −1 or x > 1,
and
f (x) is decreasing ⇔ f 0 (x) < 0 ⇔ −1 < 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. f 00 (x) a line with a positve slope, we see that:
f (x) is concave up ⇔ f 00 (x) > 0 ⇔ x > 0,
and
f (x) is concave down ⇔ f 00 (x) < 0 ⇔ x < 0.
(c) Sketch the graph of f (x). Clearly label any extreme values and inflection points.
Solution: Marking scheme: 2pts. 1pt clearly labeling the extreme values and 1pt
for labelling the inflection point.
The intercepts are given by:
x-intercepts: (−1, 0) and (2, 0). y-intercept: (0, −2)
The critical numbers are x = −1 and x = 1, and the possible inflection point is at
x = 0. 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)
(−∞, −1) (−1, 0) (0, 1) (1, 2) (2, ∞)
+
+
+
+
+
+
+
Consequently, there must be a local max at (−1, 0), local min at (1, −4), and an
inflection point when (0, −2).
The graph should look something like this:
104/184 Quiz #4
November 19
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) concave down?
Answer:
(−∞, 0.2) or (2.5, ∞)
Solution: f (x) concave down ⇔ f 00 (x) < 0 ⇔ f 0 (x) is decreasing.
(b) Where does f (x) have a local minimum?
Answer: x = 2
Solution: When f 0 (x) changes from ‘-’ to ‘+’.
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) When is the function g(x) = xe6x increasing?
Answer: x > −1/6
Solution: Marking scheme: 1pt for a correct first line of differentiation. 1pt for a
correct final answer We’ll need the derivative:
g(x) = xe6x
g 0 (x) = e6x + xe6x · 6
g 0 (x) = e6x (1 + 6x)
Therefore g 0 (x) > 0 if and only if x > −1/6.
(b) Suppose that f (x) is a function with f 0 (π) = 0 and f 00 (x) =
x2 + 9
. Determine
2x · (6 − 2x)
whether there is a local maximum or minimum value at x = π.
Answer: local maximum
Solution: Marking scheme: 1pt for determining that f 00 (π) < 0. 1pt for a correct
final answer.
The value f 00 (π) is given by
f 00 (π) =
π2 + 9
<0
2π · (6 − 2π)
since π > 3 implies that 6 − 2π < 0. Therefore, f (π) is a local maximum value by the
Second Derivative Test.
Long answer question — you must show your work
3. 4 marks Consider the function
f (x) = (x + 2)(x − 1)2 ,
which has derivatives
f 0 (x) = 3(x2 − 1) and f 00 (x) = 6x.
(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.
f 0 (x) a parabola open upwards with roots x = −1 and x = 1. Therefore,
f (x) is increasing ⇔ f 0 (x) > 0 ⇔ x < −1 or x > 1,
and
f (x) is decreasing ⇔ f 0 (x) < 0 ⇔ −1 < 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.
f 00 (x) a line with a positve slope, we see that:
f (x) is concave up ⇔ f 00 (x) > 0 ⇔ x > 0,
and
f (x) is concave down ⇔ f 00 (x) < 0 ⇔ x < 0.
(c) Sketch the graph of f (x). Clearly label any extreme values and inflection points.
Solution: Marking scheme: 2pts. 1pt clearly labeling the extreme values and 1pt
for labelling the inflection point.
The intercepts are given by:
x-intercepts: (−2, 0) and (1, 0). y-intercept: (0, 2)
The critical numbers are x = −1 and x = 1, and the possible inflection point is at
x = 0. 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)
(−∞, −2) (−2, −1) (−1, 0) (0, 1) (1, ∞)
+
+
+
+
+
+
+
+
+
Consequently, there must be a local max at (−1, 4), local min at (1, 0), and an inflection point when (0, 2).
The graph should look something like this: