a. Graph f x = x 2 − 3 ; x ≤ 0. (See Example 7) b. From the graph of f , is f a one-to-one function? c. Write the domain of f in interval notation. d. Write the range of f in interval notation. e. Write an equation for f − 1 x . f. Graph y = f x and y = f − 1 x on the same coordinate system . g. Write the domain of f − 1 in interval notation. h. Write the range of f − 1 in interval notation.
a. Graph f x = x 2 − 3 ; x ≤ 0. (See Example 7) b. From the graph of f , is f a one-to-one function? c. Write the domain of f in interval notation. d. Write the range of f in interval notation. e. Write an equation for f − 1 x . f. Graph y = f x and y = f − 1 x on the same coordinate system . g. Write the domain of f − 1 in interval notation. h. Write the range of f − 1 in interval notation.
Solution Summary: The author analyzes the graph of the function f(x) and determines whether it is a one-to-one function.
b. From the graph of
f
, is
f
a one-to-one function?
c. Write the domain of
f
in interval notation.
d. Write the range of
f
in interval notation.
e. Write an equation for
f
−
1
x
.
f. Graph
y
=
f
x
and
y
=
f
−
1
x
on the same coordinate system.
g. Write the domain of
f
−
1
in interval notation.
h. Write the range of
f
−
1
in interval notation.
System that uses coordinates to uniquely determine the position of points. The most common coordinate system is the Cartesian system, where points are given by distance along a horizontal x-axis and vertical y-axis from the origin. A polar coordinate system locates a point by its direction relative to a reference direction and its distance from a given point. In three dimensions, it leads to cylindrical and spherical coordinates.
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
Evaluate the integral using integration by parts.
Sx² cos
(9x) dx
A Problem Solving Approach To Mathematics For Elementary School Teachers (13th Edition)
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