No. If we shift any function f one unit left, we get a new function h(x)
= f(x+1). Now, there is no reason to believe that, in general, h(-x) = f(-x+1)
would be equal to -h(x) = -f(x+1). Indeed, if we let f(x) = x3,
then f is an odd function, but h(x) = f(x+1) = (x+1)3 is certainly
not odd.
(4)
No. Consider u(y) = [Ö(y2+1)]. Letting
h(y) = u(y+1), we have that
h(y) =
________
Ö(y+1)2 +
1
=
_________
Ö(y2 + 2y
+ 2
,
which is easily seen not to be an even function.
(5)
Let f be an even function an let h(x) = f(x)+1. Then h(-x) = f(-x) + 1 =
f(x) + 1 = h(x). Hence, h is an even function as well. Even functions are
symmetric about the vertical axis. Consequently, a horizontal translation
of any (nonconstant) even function destroys this symmetry and results in
a function which is no longer even. However, any vertical translation of
an even function results in another even function. In short, we can say the
eveness of a function is a property which is preserved under vertical
translations but not under horizontal translations. QUESTION: What can
we say about odd functions and translations?
(6)
No. Any nonzero vertical translation of an odd function will destroy its
diagonal symmetry.
(7)
Let f(x) = 0 . Clearly, then, f(-x) = 0 = f(x) and f(-x) = 0 = -f(x); hence,
f is both even and odd. Since a = -a is only true if a = 0, it follows that
no function of the form f(x) = a can be odd if a
¹ 0.
(8)
An even function must be symmetric with respect to the vertical axis. Thus,
an even function will fail the horizontal line test (unless its graph consists
of a single point on the vertical axis).
(9)
Let f be an odd, invertible function, and let y = f(x). We know then that
x = f-1(y). Since f is odd, we know f(-x) = -f(x); or, equivalently,
that y = f(x) = -f(-x). Consequently, we know that -y = f(-x), which implies
that -x = f-1(-y). Thus, x = -f-1(-y), which tells
us that f-1 is odd.
(10)
The graph of h is the graph of f shifted one unit left and three units up.
Note that h(x) = [(x+1)2 - 3(x+1) + 1]+3 = x2 - x +
2.
(11)
h(x) = (-1/2)[Ö(x+4)] - 5.
(12)
Let f(x) = 1/x. We are looking for a formula of the type g(x) = af(x-h) +
v, where a, h, and v are constants. We know that f has a vertical asymptote
at x = 0. Since g has a vertical asymptote at x = -1, we know that g involves
a horizontal translation one unit left; hence, h = -1. We know that f has
a horizontal asymptote at y = 0. Since g has a horizontal asymptote at y
= -2, we know that g involves a vertical shift two units down; hence v =
-2. Thus, we so far know that g(x) = af(x+1) - 2. To find the constant a,
we choose a point on the graph of g, say (0,1). We know that 1 = g(0) = af(0+1)
- 2 = a - 2 (since f(0+1) = f(1) = 1). Thus, a = 3. The desired formula is
g(x) =
3
x+1
+ 2.
(13)
This problem is worked in much the same way as Problem 12. The answer is
h(x) = -2(x-3)3 + 1.
(14)
If f(x) = x3 - 2, then y = x3 -2. Solving for x gives
us x = (y+2)1/3.
(15)
This problem is solved in the same manner as Problem 14; the answer is
f-1(y) = 2+ y5/32.
(16) We must compute f(g(x)) and g(f(x)). Based on the
graphical information given, we can accomplish this for x=-2,-1,0,1, and
x=2. For example, we know f(g(-2)) = f(2) = 1; hence, the point (-2,1)
lies on the graph of f °g. Proceeding
in this manner, we obtain the following graphs:
QUESTION: Why can we assume that the points we plot for f(g(x)) and
g(f(x)) are connected by straight lines?
