For Exercises 104-105, we demonstrate that trigonometric functions can be approximated by polynomial functions over a given interval in the domain. Graph functions f , g , h , and k on the viewing window − 4 ≤ x ≤ 4 , − 4 ≤ y ≤ 4 . Then use a Table feature on a graphing utility to evaluate each function for the given values of x . How do functions g , h , and k compare to function f for x values farther from 0 ?
For Exercises 104-105, we demonstrate that trigonometric functions can be approximated by polynomial functions over a given interval in the domain. Graph functions f , g , h , and k on the viewing window − 4 ≤ x ≤ 4 , − 4 ≤ y ≤ 4 . Then use a Table feature on a graphing utility to evaluate each function for the given values of x . How do functions g , h , and k compare to function f for x values farther from 0 ?
Solution Summary: The author explains how to graph functions using a graphing utility on the viewing window, f,g,handk.
For Exercises 104-105, we demonstrate that trigonometric functions can be approximated by polynomial functions over a given interval in the domain.
Graph functions
f
,
g
,
h
, and
k
on the viewing window
−
4
≤
x
≤
4
,
−
4
≤
y
≤
4
. Then use a Table feature on a graphing utility to evaluate each function for the given values of
x
. How do functions
g
,
h
, and
k
compare to function
f
for
x
values farther from
0
?
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
College Algebra with Modeling & Visualization (5th Edition)
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.