Honors 2201-4
Homework 3 Solutions
Fall 2003
Section 1.8
(2, 14) The inverse relation is not a function since there are two values of y
for every value of x. The domain of f (x) is all real numbers, except 0. In
symbols x ∈ <, x 6= 0.
x=
1
y2
1
= y2
x
1
y = ± √ = f −1 (x)
x
(7, 19) The inverse relation is a function. The domain of f (x) is all real
numbers, x ∈ <. The domain of f −1 (x) is all positive numbers, x ∈ <,
x > 0.
x = 2y
log2 x = y = f −1 (x)
(29)
x = my + b
x − b = my
b
1
= f −1 (x)
y = x−
m
m
The inverse is a line with slope 1/m and intercept b/m.
(30) A linear function with slope zero is a horizontal line. Its inverse relation
will be a vertical line, and will therefore fail the vertical line test. For each
value of x, there will be infinite values of y. Therefore, the inverse relation
will not be a function.
(35) The function in the figure will not have an inverse function because there
are multiple x values that correspond to the same y values. For example,
f (−3) = f (0) = f (3) = 0.
1
(36) The function in the figure is an odd function. It is symmetric over the
origin. We also notice that it satisfies the formula condition, f (−x) = −f (x).
For example, f (−1.6) = −1 and f (1.6) = 1.
(38) An even function with more than one point cannot have an inverse function. From the formula definition, we know that if f (a) = b, then f (−a) = b.
Therefore, in the inverse function, two values will correspond to f −1 (b) = ±a.
Without a unique pairing, we won’t have a function. Also, note that due to
symmetry, an even function will fail the horizontal line test for inverse functions.
The inverse of an odd function, if one exists, will always be odd. From the
formula definition we know that if f (a) = b, then f (−a) = −b. Therefore,
using the inverse function, we see that f −1 (b) = a, and f −1 (−b) = −a. This
is the criteria for a function to be odd.
(39) The equation does not define y as
√a function of x because for each value
of x there are two values of y, y = ± 4 − x2 . The equation does not define
x as a function of y√for a similar reason. For each value of y there are two
values of x, x = ± 4 − y 2 . In order for y to be a function of x, we must
restrict y to be only positive or only negative numbers (y ≥ 0, or y ≤ 0).
Likewise, for x to be a function of y, we must restrict x to be only positive
or negative numbers (x ≥ 0, or x ≤ 0). For both of the conditions to be
true, we must specify one of four scenarios, x ≥ 0, y ≥ 0, or x ≥ 0, y ≤ 0,
or x ≤ √
0, y ≤ 0, or x ≤ 0, y ≥ 0. If x ≥ 0 and√y ≥ 0, and the function
is y = 4 − x2 , then its inverse function is x = 4 − y 2. The other three
cases follow in a similar fashion. Looking at these functions we see that y as
a function of x is the inverse of x as a function of y.
Section 1.9
(2) 9π/5 ≈ 5.654866776
A 4SD approximation is 5.655
The correct 3SD approximation is 5.65. Notice that the correct approximation is less than 0.005 smaller than the real value. The incorrect approximation (5.66) is more than 0.005 greater than the real value. This provides one
indication that the correct approximation is better.
2
Problem 2 Graph of f(x)
6
y
4
2
0
-3
-2
-1
0
x
1
2
3
y
Problem 2 Graph of the inverse relation for f(x)
9
7
5
3
1
-1
-3
-5
-7
-9
0
1
2
3
x
Problem 7 Graph of f(x)
8
y
6
4
2
0
-3
-2
-1
0
x
3
1
2
3
Problem 7 Graph of the inverse relation for f(x)
y
2
0
-2
0
1
2
3
4
x
5
6
7
8
Problem 39 Graph of y as a function of x
y
2
0
-2
-1
0
x
1
2
Problem 39 y and x as functions of one another
y
2
0
0
1
x
4
2