Functions and their curves 189

y

y 52 x 11

4 y^5 x

2

234

1

21 0 1

21

x

y5 2x 2 12

Figure 18.30

y

y 5 x

y 5 x

y 5 x^2

4

(^123) x
2
0
Œ„
Figure 18.31
f−^1 (x)=
√
xforx>0isshowninFig.18.31and,again,
f−^1 (x)isseentobeareflectionoff(x)intheliney=x.
It is noted from the latter example, that not all func-
tions have an inverse. An inverse, however, can be
determined if the range is restricted.
Problem 5. Determine the inverse for each of the
following functions:
(a) f(x)=x−1(b)f(x)=x^2 − 4 (x> 0 )
(c) f(x)=x^2 + 1
(a) Ify=f(x),theny=x− 1
Transposing forxgivesx=y+ 1
Interchangingxandygivesy=x+ 1
Hence iff(x)=x−1, thenf−^1 (x)=x+ 1
(b) Ify=f(x),theny=x^2 − 4 (x> 0 )
Transposing forxgivesx=
√
y+ 4
Interchangingxandygivesy=
√
x+ 4
Hence iff(x)=x^2 − 4 (x> 0 )then
f−^1 (x)=
√
x+4ifx>− 4
(c) Ify=f(x),theny=x^2 + 1
Transposing forxgivesx=
√
y− 1
Interchangingxandygivesy=
√
x−1,whichhas
two values.
Hence there is no inverse off(x)=x^2 + 1 ,since
the domain off(x)is not restricted.
Inverse trigonometric functions
If y=sinx,thenx istheanglewhosesineisy.
Inverse trigonometrical functions are denoted by pre-
fixing the function with ‘arc’ or, more commonly,−^1.
Hence transposingy=sinx forx givesx=sin−^1 y.
Interchangingxandygives the inversey=sin−^1 x.
Similarly, y=cos−^1 x, y=tan−^1 x, y=sec−^1 x,
y=cosec−^1 xandy=cot−^1 xare all inverse trigono-
metric functions. The angle is always expressed in
radians.
Inverse trigonometric functions are periodic so it is
necessarytospecifythesmallestorprincipalvalueofthe
angle. For sin−^1 x,tan−^1 x,cosec−^1 xand cot−^1 x,the
principal value is in the range−
π
2
<y<
π
2
.Forcos−^1 x
and sec−^1 xthe principal value is in the range 0<y<π.
Graphs of the six inverse trigonometric functions are
shown in Fig. 33.1, page 335.
Problem 6. Determine the principal values of
(a) arcsin 0.5 (b) arctan(− 1 )
(c) arccos
(
−
√
3
2
)
(d) arccosec(
√
2 )
Using a calculator,
(a)arcsin0. 5 ≡sin−^10. 5 = 30 ◦

π
6
rador0.5236rad
(b)arctan(− 1 )≡tan−^1 (− 1 )=− 45 ◦
=−
π
4
rador−0.7854rad