UNIT 5 QUIZ SOLUTIONS
(100 points)
1. (10 points) Find the difference quotient of these functions:
a) f(x)= -3x-5
Solution:
f(x) = -3x-5
f(x+ ) = -3x -3 - 5
f(x+ ) – f(x) = (-3x -3
–
–
=
b) f(x)=
– 5) +3x + 5
=
= -3
- 4x
Solution:
f(x)= - 4x
f(x+
=
f(x+ ) – f(x) =
–
- 4x -4
- 4x -4
=
-
= 2x-4
2. (10 points) Find the derivative of these functions using the difference quotient.
a)
√
Solution:
f
√
f(x+
=√
f(x+ ) – f(x)= √
–
=
–
b)
Solution:
-√
√
√
=
√
f(x+
) – f(x) =
–
-
=
=
–
=
3. (10 points) Find the limit:
a)
Solution:
= (10-3)(4+3)
= 49
b)
Solution:
=
= -2
4. (10 points) Find the derivative of these functions by applying the Power Rule:
a) f(x)=
Solution:
f(x)=
let u = 2x-7
so f’(x) = 3(2x-7)2 [
= 3(2)(2x-7)2
= 6(2x-7)2
]
b) s(t) = √
Solution:
s(t) =
let u =
so s’(t) =
[
so s’(t) =
(4t +5)
s’(t)=
√
]
5. (10 points) Find the derivative of these functions by applying the Product or
Quotient Rule:
a) f(x) =(
Solution:
Using the product rule:
f(x) =(
=(
=4x4 +
=(10x4+
(4x+3) +
- 9 x + 6x4 - 18x -15)
(
+
-3)
-9x +
-15
c) h(t) =
Solution:
Using the quotient rule:
h(t) =
=
(
)
6. (25 points) The temperature T (in degrees Fahrenheit) of food placed in a
freezer can be modeled by:
T=
Where t is the time in hours.
Find the rates of change of T when t=1, t=3, t=5, t=10
Using the quotient rule:
= (
)=
=
at t=1,
=
at t=3
=
at t=5
=
at t=5
=
7. The current flowing through the capacitor shown below is given by i(t) =
If C=500 μF, and v(t) = 250 sin(200
a) Find the current, i(t)
Solution:
i(t) =
= 500x10-6 (250 sin(200
= 500x10-6 (250 (200 cos(200
i(t)=25 cos(200
A
b) Find the power (p) = v(t) i(t), and its maximum value pmax.
Solution:
p(t) = v(t) i(t)
= i(t)=25 cos(200
= 6250 cos(200
W
now we simplify using the trig identity: sin2x=2sinx cosx
p(t) = 3125 sin(400
pmax = 3125
W ; the maximum is the amplitude
.