Higher Engineering Mathematics, Sixth Edition

188 Higher Engineering Mathematics

(a) A graph ofy=lnx is shown in Fig. 18.29(a)

and the curve is neither symmetrical about the

y-axis nor symmetrical about the originand is thus

neither even nor odd.

(b) A graph ofy=xin the range−πtoπis shown in

Fig. 18.29(b) and is symmetrical about the origin

and is thus anodd function.

(a)

y

x

yIn x

0.5

1.0

0.5

02341

(b)

y

x

yx

0 





 2  2 

Figure 18.29

Now try the following exercise

Exercise 78 Further problems on even and

odd functions

In Problems 1 and 2 determine whether the given

functions are even, odd or neither even nor odd.

1. (a)x^4 (b) tan3x (c) 2e^3 t (d) sin^2 x
   [
   (a)even (b)odd
   (c)neither (d)even

]

2. (a) 5t^3 (b) ex+e−x (c)

cosθ

θ

(d) ex

[

(a)odd (b)even

(c)odd (d)neither

]

3. State whether the following functions, which
   are periodic of period 2π, are even or odd:

(a) f(θ )=

{

θ, when−π≤θ≤ 0

−θ, when 0≤θ≤π

(b)f(x)=

⎧

⎨

⎩

x, when−

π

2

≤x≤

π

2

0 , when

π

2

≤x≤

3 π

2

[(a) even (b) odd]

18.6 Inverse functions

Ifyis a function ofx, the graph ofyagainstxcan be

used to findxwhen any value ofyis given. Thus the

graph also expresses thatxis a function ofy.Twosuch

functions are calledinverse functions.

In general, given a functiony=f(x), its inverse may

be obtained by interchanging the roles ofxandyand

then transposing fory. The inverse function is denoted

byy=f−^1 (x).

For example, ify= 2 x+1, the inverse is obtained by

(i) transposing forx,i.e.x=

y− 1

2

=

y

2

−

1

2

and

(ii) interchanging x and y, giving the inverse as

y=

x

2

−

1

2

Thus iff(x)= 2 x+1, thenf−^1 (x)=

x

2

−

1

2

A graph of f(x)= 2 x+1 and its inversef−^1 (x)=

x

2

−

1

2

is shown in Fig. 18.30 andf−^1 (x)is seen to be

a reflection off(x)in the liney=x.

Similarly, ify=x^2 , the inverse is obtained by

(i) transposing forx,i.e.x=±

√

yand

(ii) interchanging x and y, giving the inverse

y=±

√

x.

Hence the inverse has two values for every value ofx.

Thus f(x)=x^2 does not have a single inverse. In

such a case the domain of the original function may

be restricted toy=x^2 forx>0. Thus the inverse is

theny=+

√

x.Agraphoff(x)=x^2 and its inverse