HW 2C - Limit Laws: Recommended review problems: #4,9,14,15,17
1. (1 pts)
Evaluate the limit:
lim𝑥→6 2 =
2. (1 pts)
Evaluate the limit:
lim𝑥→−8 𝑥 =
3. (1 pts)
Evaluate the limit:
lim𝑥→ 2 7𝑥 − 4 =
4. (1 pts)
Evaluate the limit
3(𝑦 2 − 1)
lim𝑦→3
6𝑦 2 (𝑦 − 1)3
5. (1 pts)
Evaluate the limit
lim𝑥→8 (7𝑥 2 + 6)(8𝑥 + 4)
6. (1 pts)
Given lim𝑥→8 𝑓(𝑥) = 3 and lim𝑥→8 𝑔(𝑥) = −5 , evaluate:
lim𝑥→8 (−8𝑓(𝑥) + 3𝑔(𝑥)) =
7. (1 pts)
Evaluate the limit: lim𝑥→−9
8. (1 pts)
Evaluate the limit
lim𝑥→6 6(3𝑥 + 6)3
𝑥 2 +8𝑥−9
𝑥+9
9. (1 pts)
Evaluate the limit: lim𝑥→0
√5𝑥+64−8
𝑥
10. (1 pts)
Evaluate the limit
𝑥 2 + 2𝑥 + 1
lim𝑥→1
𝑥+1
11. (1 pts)
Evaluate the limit
64 − 𝑏
lim𝑏→64
8 − √𝑏
12. (1 pts)
Find the limit:
lim𝑥→144
√𝑥 − 12
𝑥 − 144
Give an exact answer or keep at least three decimal places.
13. (1 pts)
Evaluate the limit:
lim𝑥→−9
−6𝑥 − 54
=
𝑥 2 + 16𝑥 + 63
14. (1 pts)
Evaluate the limit: lim𝑥→−3
1 1
−
𝑥 −3
𝑥+3
. Enter your answer as a reduced fraction.
15. (1 pts)
Find the following limit:
lim𝑥→10
𝑥 3 − 5𝑥 2 − 52𝑥 + 20
𝑥 − 10
Answer:
16. (1 pts)
Evaluate the limit: limℎ→0
(−3+ℎ)2 −9
ℎ
17. (1 pts)
1
Evaluate: lim𝑥→0 (2𝑥 3 sin (𝑥) + 9)
18. (1 pts)
Estimate the limit numerically or state that the limit does not exist:
lim𝑥→0
sin(9𝑥)
𝑥
Give your answer to at least three decimal places
19. (1 pts)
Evaluate the limit, if it exists: lim𝑥→−5
𝑥 4 −625
𝑥 3 +125
20. (1 pts)
Evaluate the limit: lim𝑥→2
𝑥−2
√𝑥 2 +12−4
. Enter your answer as a reduced fraction.
21. (1 pts)
Find the limit lim𝑥→1
2𝑥 2 −7𝑥+5
5𝑥 2 −7𝑥+2
22. (1 pts)
2
if 𝑥 ≤ 4
Let 𝑓(𝑥) = {6 − 𝑥 − 𝑥
2𝑥 − 7 if 𝑥 > 4
Calculate the following limits. Enter "DNE" if the limit does not exist.
lim𝑥→4− 𝑓(𝑥) =
lim𝑥→4+ 𝑓(𝑥) =
lim𝑥→4 𝑓(𝑥) =
23. (1 pts)
√−5 − 𝑥 + 3 if 𝑥 < −6
Let 𝑓(𝑥) = {
3 if 𝑥 = −6
2𝑥 + 16 if 𝑥 > −6
Calculate the following limits. Enter "DNE" if the limit does not exist.
lim𝑥→−6− 𝑓(𝑥) =
lim𝑥→−6+ 𝑓(𝑥) =
lim𝑥→−6 𝑓(𝑥) =
+++++++++++++++
Key - Form 1
i.
ii.
iii.
iv.
v.
vi.
vii.
viii.
ix.
x.
xi.
xii.
2
-8
10
0.055555555555556
30872
-39
-10
82944
0.3125
2
16
1
24
xiii. 3
xiv. -0.11111111111111
xv. 148
xvi. -6
xvii. 9
xviii. 9
xix. -6.6666666666667
xx. 2
xxi. -1
xxii. -14 ~ 1 ~ 𝐷𝑁𝐸
xxiii. 4 ~ 4 ~ 4
HW 2D - Formal Definition of Limit: Recommended review problems: #1,4
1. (1 pts)
What word or phrase best fits in the blank below?
"If lim𝑥→𝑎 𝑓(𝑥) = 𝐿 , then when 𝑥 is near 𝑎 the corresponding values of 𝑓(𝑥) are close
(________________) to L."
•
and possibly equal
•
and always equal
•
but slightly smaller when compared
•
but never equal
•
but slightly larger when compared
•
as they can be
2. (1 pts)
When demonstrating that lim𝑥→0 3𝑥 + 5 = 5 with 𝜖 = 0.4 , which of the following 𝛿 -values
suffices? [There may be MORE THAN ONE correct answer, so select all that apply]
•
𝛿 = 0.26666666666667
•
𝛿 = 0.017777777777778
•
𝛿 = 0.044444444444444
•
𝛿 = 0.13333333333333
3. (1 pts)
1
In formally proving that lim𝑥→−8 (9 𝑥 − 2) = −
function of 𝜖 .
26
9
, let 𝜖 > 0 be arbitrary. Determine 𝛿 as a
Note: in this case 𝛿 will be a function of 𝜖 . You will need to write the word epsilon for
$\displaystyle\ \epsilon\$ in the answerbox
𝛿=
4. (1 pts)
In formally proving that lim𝑥→4 (𝑥 2 − 8𝑥) = −16 , let 𝜖 > 0 be arbitrary. Determine 𝛿 as a
function of 𝜖 .
Note: in this case 𝛿 will be a function of 𝜖 . You will need to write the word epsilon for
$\displaystyle\ \epsilon\$ in the answerbox
𝛿=
5. (1 pts)
In formally proving that lim𝑥→4 (𝑥 2 + 𝑥) = 20 , let 𝜖 > 0 be arbitrary. Choose
𝜖
$\displaystyle\delta\ =\$ min (𝑚 , 1) . Determine 𝑚 .
$\displaystyle{m}=\$
6. (1 pts)
In formally proving that lim𝑥→1 (𝑥 3 − 𝑥 2 + 3𝑥) = 3 , let 𝜖 > 0 be arbitrary. Choose
𝜖
$\displaystyle\delta=\$ min (𝑚 , 1) . Determine 𝑚 .
𝑚=
+++++++++++++++
Key - Form 1
i.
ii.
and possibly equal
𝛿 = 0.044444444444444
𝛿 = 0.13333333333333
𝛿 = 0.017777777777778
iii. 9 ⋅ 𝜖
iv. √𝜖
v. 10
vi. 7
HW 2E – Continuity : Recommended Review Problems: 4,11,20
1. (1 pts)
Suppose that 𝑓(𝑥) is a continuous function with 𝑓(−4) = −7 and 𝑓(0) = 7 . Determine
which choice best describes the following statement.
" 𝑓(𝑥) = 0 for some 𝑥 in the interval [-4, 0]"
•
Always false
•
Sometimes true and sometimes false
•
Always true
2. (1 pts)
Suppose that 𝑓(𝑥) is a function with 𝑓(4) = −2 and 𝑓(8) = 2 . Determine which choice best
describes the following statement.
" 𝑓(𝑥) = 0 for some 𝑥 in the interval [4, 8]"
•
Always false
•
Always true
•
Sometimes true and sometimes false
3. (1 pts)
5𝑥 − 4
Let 𝑓(𝑥) = {
−5𝑥 + 𝑏
if 𝑥 ≤ 10
if 𝑥 > 10
If 𝑓(𝑥) is a function which is continuous everywhere, then we must have
𝑏=
4. (1 pts)
𝑚𝑥 − 17
Let 𝑓(𝑥) = { 2
𝑥 + 10𝑥 − 7
if 𝑥 < −10
if 𝑥 ≥ −10
If 𝑓(𝑥) is a function which is continuous everywhere, then we must have
𝑚=
5. (1 pts)
Find 𝑐 such that the function
2
𝑓(𝑥) = { 𝑥 − 3 𝑥 ≤ 𝑐
10𝑥 − 28 𝑥 > 𝑐
is continuous everywhere.
𝑐=
6. (1 pts)
A function 𝑓(𝑥) is said to have a removable discontinuity at 𝑥 = 𝑎 if:
1. 𝑓 is either not defined or not continuous at 𝑥 = 𝑎 .
2. 𝑓(𝑎) could either be defined or redefined so that the new function IS continuous at 𝑥 = 𝑎
.
2𝑥 2 +6𝑥−80
Let 𝑓(𝑥) =
𝑥−5
Show that 𝑓(𝑥) has a removable discontinuity at 𝑥 = 5 and determine what value for 𝑓(5)
would make 𝑓(𝑥) continuous at 𝑥 = 5 .
Must define 𝑓(5) = .
7. (1 pts)
A function 𝑓(𝑥) is said to have a jump discontinuity at 𝑥 = 𝑎 if:
1. lim𝑥→𝑎− 𝑓(𝑥) exists.
2. lim𝑥→𝑎+ 𝑓(𝑥) exists.
3. The left and right limits are not equal.
if 𝑥 < 3
if 𝑥 ≥ 3
𝑥+7
Show that 𝑓(𝑥) has a jump discontinuity at 𝑥 = 3 by calculating the limits from the left and
right at 𝑥 = 3 .
lim𝑥→3− 𝑓(𝑥) =
lim𝑥→3+ 𝑓(𝑥) =
Now for fun, try to graph 𝑓(𝑥) .
Let 𝑓(𝑥) = {
4𝑥 − 6
4
8. (1 pts)
Which of the following graphs show a jump discontinuity? (Select all that apply)
•
•
•
9. (1 pts)
The graph below is the function 𝑓(𝑥)
Determine which one of the following explains why continuity is violated at 𝑥 = −1 .
•
𝑓(𝑎) is undefined.
•
lim𝑥→𝑎 𝑓(𝑥) and 𝑓(𝑎) exist but are not equal.
•
lim𝑥→𝑎 𝑓(𝑥) does not exist.
10. (1 pts)
The graph below is the function 𝑓(𝑥)
Determine which of the following explains why continuity is violated at 𝑥 = 2 .
•
lim𝑥→𝑎 𝑓(𝑥) does not exist.
•
𝑓(𝑎) is not defined.
•
lim𝑥→𝑎 𝑓(𝑥) and 𝑓(𝑎) exist but are unequal.
11. (1 pts)
The graph below is the function 𝑓(𝑥)
Determine which of the following explains why continuity is violated at 𝑥 = 2 .
•
𝑓(𝑎) is not defined.
•
lim𝑥→𝑎 𝑓(𝑥) and 𝑓(𝑎) exist but are unequal.
•
lim𝑥→𝑎 𝑓(𝑥) does not exist.
12. (1 pts)
The graph below is the function 𝑓(𝑥)
Select all statements below that you agree with.
Note: You may be checking more than one box. No partial credit.
•
𝑓(2) is defined.
•
lim𝑥→2 𝑓(𝑥) exists.
•
lim𝑥→2 𝑓(𝑥) = 𝑓(2) .
•
The function is continuous at x = 2.
•
The function is not continuous at x = 2.
13. (1 pts)
The graph below is the function 𝑓(𝑥)
Select all statements below that you agree with.
Note: You may be checking more than one box. No partial credit.
•
𝑓(2) is defined.
•
lim𝑥→2 𝑓(𝑥) exists.
•
lim𝑥→2 𝑓(𝑥) = 𝑓(2) .
•
The function is continuous at x = 2.
•
The function is not continuous at x = 2.
14. (1 pts)
The graph below is the function 𝑓(𝑥)
Determine which one of the following rules for continuity is first violated at 𝑎 = −1 .
•
𝑓(𝑎) is defined.
•
lim𝑥→𝑎 𝑓(𝑥) exists.
•
lim𝑥→𝑎 𝑓(𝑥) = 𝑓(𝑎) .
15. (1 pts)
The graph below is the function 𝑓(𝑥)
Determine which one of the following rules for continuity is violated first at 𝑥 = 2 .
•
𝑓(𝑎) is defined.
•
lim𝑥→𝑎 𝑓(𝑥) exists.
•
lim𝑥→𝑎 𝑓(𝑥) = 𝑓(𝑎) .
16. (1 pts)
The graph below is the function 𝑓(𝑥)
Determine which one of the following rules for continuity is violated first at 𝑥 = 2 .
•
𝑓(𝑎) is defined.
•
lim𝑥→𝑎 𝑓(𝑥) exists.
•
lim𝑥→𝑎 𝑓(𝑥) = 𝑓(𝑎) .
17. (1 pts)
𝑚𝑥 − 8 if 𝑥 < −3
Let 𝑓(𝑥) = { 2
𝑥 + 9𝑥 − 5 if 𝑥 ≥ −3
If 𝑓(𝑥) is a function which is continuous everywhere, then we must have
𝑚=
Now for fun, try to graph 𝑓(𝑥) .
18. (1 pts)
Suppose that 𝑓(𝑥) is a continuous function with 𝑓(−1) = −1 and 𝑓(3) = 1 . Determine
which choice best describes the following statement.
"For any 𝑦 in the interval [-1,1], 𝑓(𝑥) = 𝑦 for some 𝑥 in the interval [-1, 3]"
•
Always false
•
Sometimes true and sometimes false
•
Always true
19. (1 pts)
Suppose that 𝑓(𝑥) is a continuous function with 𝑓(4) = −8 and 𝑓(8) = 8 . Determine which
choice best describes the following statement.
"For some 𝑦 in the interval [-8,8], 𝑓(𝑥) = 𝑦 for all 𝑥 in the interval [4, 8]"
•
Sometimes true and sometimes false
•
Always true
•
Always false
20. (1 pts)
Find the values of 𝑚 and 𝑏 that make 𝑓 continuous everywhere:
48 + 2𝑥 − 𝑥 2
if 𝑥 < −6
𝑥+6
𝑓(𝑥) = {
𝑚𝑥 + 𝑏
if 𝑥 ∈ [−6,4]
4−𝑥
4⋅2
+ 50
if 𝑥 > 4
When showing your work, be sure to include all limit calculations and a sketch of the graph.
𝑚=
𝑏=
21. (1 pts)
Given the function below, determine if the function is continuous at the point 𝑥 = 3 . If not,
indicate why.
2𝑥 − 11, 𝑥 < 3
2, 𝑥 = 3
−3𝑥 + 4, 𝑥 > 3
Continuous at 𝑥 = 3
Not continuous: 𝑓(3) does not exist
Not continuous: lim𝑥→3 𝑓(𝑥) does not exist
Not continuous: 𝑓(3) and limit exist, but are not equal
𝑓(𝑥) = {
•
•
•
•
22. (1 pts)
Given the function below, determine if the function is continuous at the point 𝑥 = −3 . If
not, indicate why.
𝑥 2 − 16
𝑥+3
Continuous at 𝑥 = −3
Not continuous: 𝑓(−3) is not defined; this is a removable discontinuity
Not continuous: 𝑓(−3) is not defined; this is not a removable discontinuity
Not continuous: lim𝑥→−3 𝑓(𝑥) does not exist
Not continuous: 𝑓(−3) and limit exist, but are not equal
𝑓(𝑥) =
•
•
•
•
•
23. (1 pts)
4𝑥 + 17 if 𝑥 < −3
Let 𝑓(𝑥) = {√𝑥 + 28 if 𝑥 > −3
2 if 𝑥 = −3
Select all statements below that you agree with.
Note: You may be checking more than one box. No partial credit.
•
𝑓(−3) is defined.
•
lim𝑥→−3 𝑓(𝑥) exists.
•
lim𝑥→−3 𝑓(𝑥) = 𝑓(−3) .
•
The function is continuous at x = -3.
•
The function is not continuous at x = -3.
+++++++++++++++
Key - Form 1
i.
ii.
iii.
iv.
v.
vi.
vii.
viii.
Always true
Sometimes true and sometimes false
96
-1
5
26
6 ~ 0.4
ix. lim𝑥→𝑎 𝑓(𝑥) does not exist.
x. 𝑓(𝑎) is not defined.
xi. lim𝑥→𝑎 𝑓(𝑥) and 𝑓(𝑎) exist but are unequal.
xii.
𝑓(2) is defined.
lim𝑥→2 𝑓(𝑥) exists.
The function is not continuous at x = 2.
xiii.
𝑓(2) is defined.
The function is not continuous at x = 2.
xiv. lim𝑥→𝑎 𝑓(𝑥) exists.
xv. 𝑓(𝑎) is defined.
xvi. lim𝑥→𝑎 𝑓(𝑥) = 𝑓(𝑎) .
xvii. 5
xviii. Always true
xix. Always false
xx. 4 ~ 38
xxi. Not continuous: 𝑓(3) and limit exist, but are not equal
xxii. Not continuous: 𝑓(−3) is not defined; this is not a removable discontinuity
xxiii.
𝑓(−3) is defined.
lim𝑥→−3 𝑓(𝑥) exists.
The function is not continuous at x = -3.
HW 2F - Limits at Infinity-Horizontal Asymptote: Recommended Review Problems: 3,5,8,9
1. (1 pts)
Evaluate the following limits. If needed, enter oo for ∞ and -oo for −∞ .
(a) lim𝑥→∞ (12𝑥 2 + 27𝑥 3 ) =
(b) lim𝑥→−∞ (12𝑥 2 + 27𝑥 3 ) =
2. (1 pts)
Evaluate the following limits. If needed, enter oo for ∞ and oo for −∞ .
(a) lim𝑥→∞
√8+11𝑥2
4+4𝑥
(b) lim𝑥→−∞
=
√8+11𝑥 2
4+4𝑥
=
3. (1 pts)
Evaluate the limit
lim𝑥→∞
√11 + 9𝑥 2
9 + 9𝑥
4. (1 pts)
Evaluate the limit
lim𝑥→∞
3 + 2𝑥
3 − 10𝑥
5. (1 pts)
Evaluate the limit
lim𝑥→∞
8𝑥 + 8
− 8𝑥 + 2
6𝑥 2
6. (1 pts)
Evaluate the limit
lim𝑥→∞
2𝑥 3 − 9𝑥 2 − 8𝑥
5 − 7𝑥 − 7𝑥 3
7. (1 pts)
Evaluate the limit
lim𝑥→∞
3𝑥 2 − 3𝑥 + 4
3𝑥 + 11
8. (1 pts)
Evaluate the following limits.
(a) lim𝑥→∞
9
𝑒 𝑥 −9
(b) lim𝑥→−∞
=
9
𝑒 𝑥 −9
=
[NOTE: If needed, enter oo for ∞ and -oo for −∞ .]
9. (1 pts)
Evaluate the following limits. Enter your answer as a reduced fraction. If needed, enter oo
for ∞ and -oo for −∞ .
lim𝑥→∞ ( √𝑥 2 + 8𝑥 − 𝑥) =
10. (1 pts)
Evaluate the following limits. If needed, enter oo for ∞ and -oo for −∞ .
lim𝑥→−∞ ( 𝑥 + √𝑥 2 + 2𝑥) =
11. (1 pts)
Evaluate the following limits. If needed, enter oo for ∞ and -oo for −∞ .
lim𝑥→∞ ( √𝑥 2 + 2𝑥 − √𝑥 2 + 4𝑥) =
+++++++++++++++
Key - Form 1
i.
∞ ~ −∞
ii. 0.82915619758885 ~ -0.82915619758885
iii. 0.33333333333333
1
iv.
−5
v.
0
vi.
−7
vii.
viii.
ix.
x.
xi.
∞
0 ~ -1
4
-1
-1
2
HW 2G - Rates of Change: Recommended Review Problems: 5,11,13
1. (1 pts)
If 𝑓(𝑥) = 3𝑥 2 − 7𝑥 + 4 , find 𝑓′(0) .
Use this to find the equation of the tangent line to the parabola 𝑦 = 3𝑥 2 − 7𝑥 + 4 at the
point (0,4) . The equation of this tangent line can be written in the form 𝑦 = 𝑚𝑥 + 𝑏
where 𝑚 is:
and where 𝑏 is:
2. (1 pts)
Let 𝑠(𝑡) = 6𝑡 3 − 36𝑡 2 + 54𝑡 be the equation of motion for a particle. Find a function for the
velocity.
𝑣(𝑡) =
Where does the velocity equal zero? [Hint: factor out the GCF. ]
𝑡 = and 𝑡 =
Find a function for the acceleration of the particle.
𝑎(𝑡) =
3. (1 pts)
Suppose that the position of a particle is given by 𝑠 = 𝑓(𝑡) = 4𝑡 3 + 3𝑡 + 9 .
(a) Find the velocity at time 𝑡 .
𝑣(𝑡) =
𝑚
𝑠
(b) Find the velocity at time 𝑡 = 3 seconds.
𝑚
𝑠
(c) Find the acceleration at time 𝑡 .
𝑚
𝑎(𝑡) = 𝑠2
(d) Find the acceleration at time 𝑡 = 3 seconds.
𝑚
𝑠2
4. (1 pts)
A particle moves along a straight line and its position at time 𝑡 is given by 𝑠(𝑡) = 2𝑡 3 −
30𝑡 2 + 126𝑡 where s is measured in feet and t in seconds.
Find the velocity (in ft/sec) of the particle at time 𝑡 = 0 :
The particle stops moving (i.e. is in a rest) twice,
first when 𝑡 = ,
and again when 𝑡 =
What is the position of the particle at time 20 ?
Finally, what is the TOTAL distance the particle travels between time 0 and time 20 ?
5. (1 pts)
Use the graph of 𝑓(𝑥) above to estimate the value of 𝑓′(2) to one decimal place.
𝑓′(2) =
6. (1 pts)
Find an equation of the tangent line to the parabola 𝑦 = 𝑥 2 at the point 𝑃(1.2,1.44) .
Start by calculating the slope of the secant line 𝑃𝑄 for different points 𝑄 that get closer to
the point 𝑃(1.2,1.44) . Fill in this table similar to those in this example by calculating the
slope of the tangent lines through 𝑃 and the point 𝑄(𝑥, 𝑥 2 ) for the values of 𝑥 provided in
the table.
𝑚𝑃𝑄
𝑥
2
1.5
1.3
1.21
1.201
1.2001
Based on the values in this table, it appears the slope of the tangent line should be 𝑚 =
Now that you have the slope 𝑚 of the tangent line and a point 𝑃(1.2,1.44) on the tangent
line, you can find the equation of the line:
tangent line is 𝑦 =
7. (1 pts)
1
2
2
The point 𝑃 (5 , 10) lies on the curve 𝑦 = 𝑥 . If 𝑄 is the point (𝑥, 𝑥) , find the slope of the
secant line 𝑃𝑄 for the following values of 𝑥 .
If 𝑥 = 0.3 , the slope of 𝑃𝑄 is:
and if 𝑥 = 0.21 , the slope of 𝑃𝑄 is:
and if 𝑥 = 0.1 , the slope of 𝑃𝑄 is:
and if 𝑥 = 0.19 , the slope of 𝑃𝑄 is:
Based on the above results, guess the slope of the tangent line to the curve at 𝑃(0.2,10) .
8. (1 pts)
Estimate the instantaneous rate of change at 𝑥 = 3
Your estimate needs to be within 10% of the exact answer.
9. (1 pts)
For each slope below, enter the value correct to four decimal places.
𝑓(𝑥+ℎ)−𝑓(𝑥)
Let 𝑓(𝑥) = 8.5𝑥 2 − 7.9𝑥 . Using the formula, 𝑚 =
, estimate the slope of the
ℎ
tangent line on the graph of 𝑓(𝑥) at 𝑥 = 9 for the following values of ℎ :
ℎ =1: 𝑚 =
ℎ = 0.5 : 𝑚 =
ℎ = 0.1 : 𝑚 =
ℎ = 0.01 : 𝑚 =
ℎ = 0.001 : 𝑚 =
10. (1 pts)
Use the limit definition of the derivative to find the slope of the tangent line to the curve
𝑓(𝑥) = 7𝑥 2 at 𝑥 = 2
11. (1 pts)
Let 𝑓(𝑥) be the function 8𝑥 2 − 10𝑥 + 10 . Then the quotient
𝑓(3+ℎ)−𝑓(3)
can be simplified to 𝑎ℎ + 𝑏 for:
ℎ
𝑎=
and
𝑏=
12. (1 pts)
Let 𝑓(𝑥) = √𝑥 + 4 . Calculate the difference quotient
𝑓(5+ℎ)−𝑓(5)
for
ℎ
ℎ = .1
ℎ = .01
ℎ = −.01
ℎ = −.1
If someone now told you that the derivative (slope of the tangent line to the graph) of 𝑓(𝑥)
1
at 𝑥 = 5 was 𝑛 for some integer 𝑛 what would you expect 𝑛 to be?
𝑛=
13. (1 pts)
Let 𝑓(𝑥) = √50 − 𝑥 . Compute 𝑓′(1) using the limit definition
𝑓′(1) =
Find an equation of the tangent line at 𝑥 = 1
𝑦=
14. (1 pts)
q
2.2 2.4 2.6 2.8 3
h(q) 372 512 705 971 1337
Estimate ℎ′(2.6) using the table above.
ℎ′(2.6) ≈
+++++++++++++++
Key - Form 1
i.
ii.
iii.
iv.
v.
vi.
vii.
viii.
ix.
x.
xi.
xii.
-7 ~ -7 ~ 4
18𝑡 2 − 72𝑡 + 54 ~ 1 or 3 ~ 1 or 3 ~ 36𝑡 − 72
12𝑡 2 + 3 ~ 111 ~ 24𝑡 ~ 72
126 ~ 3 ~ 7 ~ 6520 ~ 6648
-2
𝑚 = 2.4 ~ equation of tangent line is 𝑦 = 2.4𝑥 − 1.44
-33.333333333333 ~ -47.619047619048 ~ -100 ~ -52.631578947368 ~ -50
6.8
153.6 ~ 149.35 ~ 145.95 ~ 145.18499999998 ~ 145.10850000022
28
8 ~ 38
0.16620625799671 ~ 0.16662039607267 ~ 0.16671298870099 ~
0.16713221964741 ~ 6
1
1
99
xiii. − 14 ~ − 14 𝑥 + 14
xiv. near 1127.6587977589
HW 2H - The Derivative as a Function: Recommended Review Problems: 2,6,8,14
1. (1 pts)
Objective [1.10]
Could the function in bottom plot be the derivative of the function in the top plot?
•
No
•
Yes
The plot of the FUNCTION (below)
The plot of its DERIVATIVE (below)
2. (1 pts)
The function graphed above has:
Positive derivative on the interval(s)
Negative derivative on the interval(s)
3. (1 pts)
This is the graph of a function.
Choose the graph of its derivative from among the following functions.
4. (1 pts)
This is the graph of a function.
Choose the graph of its derivative from among the following functions.
5. (1 pts)
Given 𝑓(𝑥) = 6 − 3𝑥 2 , find 𝑓′(𝑥) using the limit definition of the derivative.
𝑓′(𝑥) =
6. (1 pts)
5
Given 𝑓(𝑥) = 𝑥 , find 𝑓′(𝑥) using the limit definition of the derivative.
𝑓′(𝑥) =
7. (1 pts)
Given 𝑓(𝑥) = 6√𝑥 + 6 , find 𝑓′(𝑥) using the limit definition of the derivative.
𝑓′(𝑥) =
8. (1 pts)
Given 𝑓(𝑥) = 5𝑥 2 + 9𝑥 + 4 , find 𝑓′(𝑥) using the limit definition of the derivative.
𝑓′(𝑥) =
Note: You must show your work, and you must use the limit definition for the derivative for
full credit.
9. (1 pts)
Given 𝑓(𝑥) = 4 − 2𝑥 2 , find 𝑓′(𝑥) using the limit definition of the derivative.
𝑓′(𝑥) =
10. (1 pts)
4
Given 𝑓(𝑥) = , find 𝑓′(𝑥) using the limit definition of the derivative.
𝑥
𝑓′(𝑥) =
11. (1 pts)
Select all the points at which the graph above is not differentiable
•
-4
•
-3
•
-2
•
-1
•
0
•
1
•
2
•
3
•
4
12. (1 pts)
Let
𝑓(𝑥) = 3𝑥 3 + 4𝑥 + 4
Use the limit definition of the derivative to calculate the derivative of 𝑓 :
𝑓′(𝑥) = .
Use the same formula from above to calculate the derivative of this new function (i.e. the
second derivative of 𝑓 ):
𝑓″(𝑥) = .
13. (1 pts)
The monthly salary S( t ) in dollars for a dock worker shown above has the following cubic
model.
𝑆(𝑡) = 0.185𝑡 3 − 7.95𝑡 2 + 101.2𝑡 + 991 where t is the number of years after 1980
Interpret S( 2000 ) = 1315
•
•
•
The monthly salary in 2000 was 1315.
The monthly salary in 2000 was $1315.
The salary in 2000 was 1315.
Interpret 𝑆′(2000) = 5.2
•
•
•
•
•
The monthly salary in 2000 was increasing by 5.2.
The monthly salary in 2000 was decreasing by 5.2 per year.
The monthly salary in 2000 was increasing by 5.2 per year.
The monthly salary in 2000 was decreasing by $5.2 per year.
The monthly salary in 2000 was increasing by $5.2 per year.
•
•
The monthly salary in 2000 was neither increasing nor decreasing.
The monthly salary in 2000 was decreasing by 5.2.
14. (1 pts)
The radius of a circular oil spill after 𝑡 minutes is given by 𝑟(𝑡) = √9𝑡 . Find the
instantaneous rate at which the radius is growing after 30 minutes.
Give your answer as a decimal approximation with at least 3 decimal places.
𝑓𝑡
min
+++++++++++++++
Key - Form 1
i.
ii.
No
(-3,-0.5) ~ (-oo,-3)U(-0.5,oo)
iii.
iv. −2 ⋅ 3 ⋅ 𝑥
v.
vi.
5
− 𝑥2
3
√𝑥+6
vii. 10 ⋅ 𝑥 + 9
viii. −2 ⋅ 2 ⋅ 𝑥
ix.
x.
xi.
4
− 𝑥2
-2
1
9 ⋅ 𝑥 2 + 4 ~ 18 ⋅ 𝑥
xii. The monthly salary in 2000 was $1315. ~ The monthly salary in 2000 was increasing
by $5.2 per year.
xiii. 0.27386127875258