38 Chapter 2Algebraic functions
To solve for x,
(i) multiply both sides of the equation by(cx 1 + 1 d): (cx 1 + 1 d)y 1 = 1 ax 1 + 1 b
(ii) expand the l.h.s.: cxy 1 + 1 dy 1 = 1 ax 1 + 1 b
(iii) subtract(ax 1 + 1 b)from both sides: cxy 1 + 1 dy 1 − 1 ax 1 − 1 b 1 = 10
(iv) collect the terms inx
1and x
0:(cy 1 − 1 a)x 1 + 1 (dy 1 − 1 b) 1 = 10
(v) subtract(dy 1 − 1 b)from both sides: (cy 1 − 1 a)x 1 = 1 −(dy 1 − 1 b)
(vi) divide both sides by(cy 1 − 1 a):
We note that step (vi) is not valid if(cy 1 − 1 a) 1 = 10 because division by zero has no
meaning. Such complications can normally be ignored.
This example demonstrates the type of algebraic manipulation routinely used in
the solution of real problems.
0 Exercises 26–29
EXAMPLE 2.10Ify 1 = 1 f(x) 1 = 1 x
21 + 11 , express xin terms of y.
We have
y 1 = 1 x
21 + 1 1,x
21 = 1 y 1 − 1 1,
yis a single-valued function of x, but xis a double-valuedfunction of y(except for
y 1 = 11 ); that is, for each real value ofy 1 > 11 there exist two real values of x(ify 1 < 11 then
xis complex).
Figure 2.4 shows how the graphs of the function and its inverse are related; graph (b)
is obtained from (a) by interchanging the xand yaxes, or by rotation around the line
x 1 = 1 y. Graph (b) also shows the double-valued nature of the inverse function.
xyfy=± − =
−1
1()
x
dy b
cy a
=− fy
−
−
=
−()
()
()
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(a )y=f(x)=x
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(b)x=f
− 1(y)=±
√
y− 1
Figure 2.4