Taxicab fares. In New York City, taxicabs change passengers $2.50 for entering a cab and then $050 for each one-fifth of a mile (or fraction thereof) traveled. (There are additional charge for slow traffic and idle times, but these are not considered in this problem.) If x presents the distance traveled in miles, then C ( x ) is the cost of the taxi fare, where C ( x ) = $ 2.50 , if x = 0 , C ( x ) = $ 3.00 , if 0 < x ≤ 0.2 , C ( x ) = $ 3.50 , if 0.2 < x ≤ 0.4 , C ( x ) = $ 4.00 , if 0.4 < x ≤ 0.6 , and so on. The graph of C is show below. (Source; New York City Taxi and Limousine Commission.) Using the graph of the taxicab fare function, find each of the following limits. if it exists. lim x → 0.25 − C ( x ) , lim x → 0.25 + C ( x ) , lim x → 0.25 C ( x )
Taxicab fares. In New York City, taxicabs change passengers $2.50 for entering a cab and then $050 for each one-fifth of a mile (or fraction thereof) traveled. (There are additional charge for slow traffic and idle times, but these are not considered in this problem.) If x presents the distance traveled in miles, then C ( x ) is the cost of the taxi fare, where C ( x ) = $ 2.50 , if x = 0 , C ( x ) = $ 3.00 , if 0 < x ≤ 0.2 , C ( x ) = $ 3.50 , if 0.2 < x ≤ 0.4 , C ( x ) = $ 4.00 , if 0.4 < x ≤ 0.6 , and so on. The graph of C is show below. (Source; New York City Taxi and Limousine Commission.) Using the graph of the taxicab fare function, find each of the following limits. if it exists. lim x → 0.25 − C ( x ) , lim x → 0.25 + C ( x ) , lim x → 0.25 C ( x )
Solution Summary: The author explains that taxicabs charge passengers 2.50 for entering a cab, and ifx=0 C(x)=3.
Taxicab fares. In New York City, taxicabs change passengers $2.50 for entering a cab and then $050 for each one-fifth of a mile (or fraction thereof) traveled. (There are additional charge for slow traffic and idle times, but these are not considered in this problem.) If x presents the distance traveled in miles, then
C
(
x
)
is the cost of the taxi fare, where
C
(
x
)
=
$
2.50
,
if
x
=
0
,
C
(
x
)
=
$
3.00
,
if
0
<
x
≤
0.2
,
C
(
x
)
=
$
3.50
,
if
0.2
<
x
≤
0.4
,
C
(
x
)
=
$
4.00
,
if
0.4
<
x
≤
0.6
,
and so on. The graph of C is show below. (Source; New York City Taxi and Limousine Commission.)
Using the graph of the taxicab fare function, find each of the following limits. if it exists.
lim
x
→
0.25
−
C
(
x
)
,
lim
x
→
0.25
+
C
(
x
)
,
lim
x
→
0.25
C
(
x
)
Given lim x-4 f (x) = 1,limx-49 (x) = 10, and lim→-4 h (x) = -7 use the limit properties
to find lim→-4
1
[2h (x) — h(x) + 7 f(x)] :
-
h(x)+7f(x)
3
O DNE
17. Suppose we know that the graph below is the graph of a solution to dy/dt = f(t).
(a) How much of the slope field can
you sketch from this information?
[Hint: Note that the differential
equation depends only on t.]
(b) What can you say about the solu-
tion with y(0) = 2? (For example,
can you sketch the graph of this so-
lution?)
y(0) = 1
y
AN
(b) Find the (instantaneous) rate of change of y at x = 5.
In the previous part, we found the average rate of change for several intervals of decreasing size starting at x = 5. The instantaneous rate of
change of fat x = 5 is the limit of the average rate of change over the interval [x, x + h] as h approaches 0. This is given by the derivative in the
following limit.
lim
h→0
-
f(x + h) − f(x)
h
The first step to find this limit is to compute f(x + h). Recall that this means replacing the input variable x with the expression x + h in the rule
defining f.
f(x + h) = (x + h)² - 5(x+ h)
=
2xh+h2_
x² + 2xh + h² 5✔
-
5
)x - 5h
Step 4
-
The second step for finding the derivative of fat x is to find the difference f(x + h) − f(x).
-
f(x + h) f(x) =
= (x²
x² + 2xh + h² -
])-
=
2x
+ h² - 5h
])x-5h) - (x² - 5x)
=
]) (2x + h - 5)
Macbook Pro
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