Show that any function F(x) that is defined for all x may be expressed as the sum of an even function and an odd
function: F(x) = E(x) + O(x).
Take E(x)= |[F(*) + F(-x)] and O(x) = \[F(x) - F(-x)].
Prove that the representation of F(x) in Problem 5.55 is unique.
If F(x) = E(x) + O(x) and F(x) = E*(x) + O*(x), then, by subtraction,
0 = e(x) + o(x)
(1)
where e(x) = E*(x) - E(x) is even and o(x) = O*(x) - O(x) is odd. Replace x by -x in (1) to obtain
0=e(x)-o(x)
(2)
But (1) and (2) together imply e(x) = o(x) = 0; that is, £*(*) = E(x) and O*(x) = O(x).
In Problems 5.57-5.63, determine whether the given function is one-one.
f(x) = mx + b for all x, where m^O.
Assume f(u)=f(v). Then, mu + b = mv + b, mu = mv, u = v. Thus, /is one-one.
/(*) = Vx for all nonnegative x.
Assume f(u)=f(v). Then, Vu = Vv. Square both sides; u = v. Thus,/is one-one.
f(x) = x
2
for all x.
/(-I) = 1 =/(!). Hence, /is not one-one.
f(x) = - for all nonzero x.
f(x) = \x\ for all x.
/(—I) = 1 = /(I). Hence, /is not one-one.
f(x) = [x] for all x.
/(O) = 0 = /(|). Hence, /is not one-one.
f(x) = x
3
for all x.
Assume /(M)=/(U). Then u
3 = v
3 . Taking cube roots (see Problem 5.84), we obtain u = v. Hence,/
is one-one.
In Problems 5.64-5.68, evaluate the expression
f(x) = x
2 -2x.
f(x + h) = (x + h)
2 - 2(x + h) = (x
2 +2xh + h
2 )-2x- 2h.
So,
f(x + h)- f(x) = [(x
2 + 2xh + h
2
)
f(x) = x + 4.
f(x + h) = x + h + 4. So, f(x + h)-f(x) = (x + h+4)-(x + 4) = h. Hence,
30
CHAPTER 5
5.54
If f(x) = x
3 -kx
2 + 2x for all x and if / is an odd function, find k.
f(l) = 3-k and /(-l)=-3-Jfc. Since / is odd, -3 - k = -(3- k) = -3 + k. Hence, -k = k,
t = n
5.55
5.56
5.57
5.58
5.59
5.60
5.61
5.62
5.63
5.64
5.65
for the given function /.
Hence,
I Assume f(u)=f(v). Then
Hence, u = v. Thus,/is one-one.