MTH131 Applied Calculus – Spring 2016
Lab 2 – Instantaneous Rates of Change and the Derivative
SOLUTION
x+2
0+2
2
=
=
= −1
x−2
0−2
−2
x+2
2+2
4
lim
=
= ⇒ DNE
x→2 x − 2
2−2
0
2
2
x −4
0 −4
−4
(b) lim
=
=
=2
x→0 x − 2
0−2
−2
1. (a) lim
x→0
x2 − 4
22 − 4
0
=
= ⇒ we must do more work.
x→2 x − 2
2−2
0
lim
We use that fact that x2 − 4 = (x + 2)(x − 2) to simplify the expression:
(x − 2)(x + 2)
x2 − 4
= lim
= lim (x + 2) = 4
x→2
x→2
x→2 x − 2
x−2
2
2
0 −4
−4
x −4
=
=
= −1
(c) lim
2
2
x→0 (x − 2)
(0 − 2)
(−2)2
lim
x2 − 4
22 − 4
0
=
= 2 ⇒ we must do more work.
2
x→2 (x − 2)
(2 − 2)2
0
lim
Using x2 − 4 = (x + 2)(x − 2) we have:
x2 − 4
(x − 2)(x + 2)
x+2
2+2
4
= lim
= lim
=
= ⇒ DNE
2
2
x→2 (x − 2)
x→2
x→2 x − 2
(x − 2)
2−2
0
lim
2. (a) We use the definition of average rate of change:
∆f
∆x
=
=
=
=
f (3) − f (0)
3−0
3
2 − 20
3−0
8−1
3−0
7
3
(b) The chart below shows the approximation of instantaneous rate on the interval [1, 1 + h]
for smaller and smaller h values: :
h
0.1
0.01
0.001
0.0001
0.00001
f (1 + h) − f (1)
21+h − 21
=
h
h
1.435
1.391
1.387
1.386
1.386
It turns out the exact answer is 2 ln 2 = 1.386294, which we will prove later in the course.
(If you don’t know what ln 2 means, we’ll be explaining that later in the course as well).
3. (a) We use the definition of average rate of change:
f (x2 ) − f (x1 )
x2 − x1
In both cases x1 = 9 and x2 = 10 (i.e., 9AM and 10AM).
i. When f (x) = 5x + 25 we have
f (x2 ) − f (x1 )
x2 − x1
5(10) + 25 − [5(9) + 25]
10 − 9
75 − 70
=
1
= 5.
=
ii. When y = f (x) = 2x2 − 35x + 217 we have
f (x2 ) − f (x1 )
x2 − x1
2(10)2 − 35(10) + 217 − [2(9)2 − 35(9) + 217]
10 − 9
67 − 64
=
1
= 3.
=
(b) We use the definition of instantaneous rate of change
lim
h→0
f (c + h) − f (c)
h
In both cases we want the instantaneous rate of change when c = 9 so the formula
becomes:
f (9 + h) − f (9)
lim
h→0
h
i. When y = f (x) = 5x + 25 we have
lim
h→0
f (9 + h) − f (9)
h
5(9 + h) + 25 − [5(9) + 25]
h
45 + 5h + 25 − [45 + 25]
= lim
h→0
h
5h
= lim
h→0 h
= lim 5
=
lim
h→0
h→0
= 5
ii. When y = f (x) = 2x2 − 35x + 217 we have
f (9 + h) − f (9)
h→0
h
lim
=
=
=
=
=
=
2(9 + h)2 − 35(9 + h) + 217 − [2(9)2 − 35(9) + 217]
h→0
h
2
2(81 + 18h + h ) − 35(9) − 35h + 217 − [2(81) − 35(9) + 217]
lim
h→0
h
2(18h + h2 ) − 35h
lim
h→0
h
36h + 2h2 − 35h
lim
h→0
h
h + 2h2
lim
h→0
h
lim 1 + 2h
lim
h→0
= 1
(c) We use the definition of the derivative
f (x + h) − f (x)
h→0
h
i. When y = f (x) = 5x + 25 we have
f 0 (x) = lim
5(x + h) + 25 − [5x + 25]
h
5x + 5h + 25 − [5x + 25]
= lim
h→0
h
5h
= lim
h→0 h
= lim 5
f 0 (x) =
lim
h→0
h→0
= 5
Using this formula we have f 0 (9) = 5, which is what we got in part (b). At 10AM
the rate of change will be f 0 (10) = 5 (note that the derivative in this case is just a
constant function).
ii. When y = f (x) = 2x2 − 35x + 217 we have
f 0 (x) =
=
=
=
=
2(x + h)2 − 35(x + h) + 217 − [2x2 − 35x + 217]
h→0
h
2
2
2(x + 2xh + h ) − 35x − 35h + 217 − [2x2 − 35x + 217]
lim
h→0
h
2
2(2xh + h ) − 35h
lim
h→0
h
4xh + 2h2 − 35h
lim
h→0
h
lim 4x + 2h − 35
lim
h→0
= 4x − 35
Using this formula we have f 0 (9) = 4(9) − 35 = 1, which is what we got in part (b).
At 10AM the rate of change will be f 0 (10) = 4(10) − 35 = 5.