Composite functions

S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]

1

Example:

Solution :

Let f(x) = 2x 4x 3, x 3

x 4, x 3

and g(x) =

2

x 3, x 4

x 2x 2, x 4

. Describe the function

(f+g)(x) and find its domain.

2x 4x 3, x 3

x 4, 3 x 4

x 4,

f(x)

x 4

,

2

x 3, x 3

x 4, 3 x 4

x 2x

g(

2, x 4

x)


2

Example:

Solution :

Let f(x) = | x 1| ; x 1

1 x; x 1

and g(x) =

x 2; x 0

x 3; x 0

then find

(f + g)x and (fg)x.

| x 1| , x 0

| x 1| , 0 x 1

1 x, x 1

f(x)

and

x 2, x 0

x 3, 0 x 1

x 3, x 1

g(x)

| x 1|

| x 1|

x 2

x 3

, x 0

f(x) g(x) , 0

x1 x ,3

x 1

x 1

3, x 1

2x 1, 1 x 0

2x 4, 0 x 1

4, x 1


3

Let f(x) = 2x 4x 3, x 3

x 4, x 3

and g(x) =

2

x 3, x 4

x 2x 2, x 4

then find the value of

(i) (f + g)(3.5) (ii) f(g(3)) (iii) (fg)(2) (iv) (f g)(4)

Example:

Solution :

(f + g)(3.5)

= f(3.5) + g(3.5)

= (3.5 – 4) + (3.5 – 3)

= 0

f(g(3))

= f((3 – 3))

= f(0)

= (0)2 – 4(0) + 3

= 3


4

(fg)(2)

= f(2)g(2)

= ((2)2 –4(2) +3) (2 –3)

= (– 1) (– 1)

= 1

(f – g)(4)

= f(4) – g(4)

= (4 – 4) – ((4)2 + 2(4) + 2)

= – 26

Let f(x) = 2x 4x 3, x 3

x 4, x 3

and g(x) =

2

x 3, x 4

x 2x 2, x 4


5

I f the functions f(x) and g(x) are defined on R R such that

f(x) 0, x rational

x, x irrational

and g(x) =

0, x irrational

x, x rational

then (f g) (x) is (A) one-one and onto (B) neither one-one nor onto (C) one-one but not onto (D) onto but not one-one

(I I TJ EE 2005)

Example:

Solution :

(f g) : R R 0 x if x rationalf g x

x 0 if x irrational

x if x rational

x if x irrational


6

Since each branch is linear function, function is one – one

Any value from y = –x(x is rational) does not match with that of y = x(x is irrational)Also when x is rational y = –x is rational

And when x is irrational y = x is irrational

Thus function takes all real values of x

Hence function is onto

x, if x rationalf g x

x, if x irrational


Example:

Solution :

I f f(x) = nn (a x ) , x > 0, n 2, n N. Then show that (fof) (x)

= x. Find also the inverse of f(x).

f(x) = (a – xn)1/ n

f(f(x)) = (a – (f(x))n)1/n

f(f(x)) = [a – ((a – xn)1/n)n]1/n

f(f(x)) = [a – ((a – xn)]1/n

f(f(x)) = [a – a + xn]1/n

f(f(x)) = (xn)1/n

f(f(x)) = x

Since f(f(x)) = x ,

f– 1 (x) = f(x)

f– 1 (x) = (a – xn)1/n


I f f(x) = 1

1 x, then find the value of f(f(x)), f(f(f(x))),

Also find the value of n times

fofof............of(x)

Example:

Solution :

f(x) = 1

1 x fof(x) =

11 f(x)

fof(x) = 1

11

1 x

= 1 x

1 x 1

= x 1

x


9

Now fofof(x) = f(f(f(x)))

= x 1

fx

= 1x 1

1x

=

xx x 1

Finding Now n times

fofof............of(x)

Now this function depends upon the value of n.

For n = 3, 6, 9, ……. or n = 3k

n times

fofof............of(x) = x

= x


10

For n = 3k + 1 ( n = 1, 4, 7, ………)

n times

fofof............of(x) = 1

1 x

For n = 3k + 2 ( n = 2, 5, 8, ………)

n times

fofof............of(x) = x 1

x


11

Example:

Solution :

Let f(x) = ax + b and g(x) = cx + d, a 0, c 0. Assume a = 1, b = 2. If (fog) (x) = (gof) (x) for all x, what can you say about c and d ?

(fog) (x) = f(g(x)) = a(cx + d) + b

(gof) (x) = f(f(x)) = c(ax + b) + d

Given that, (fog) (x) = (gof) (x) and at a = 1, b = 2

cx + d + 2 = cx + 2c + d

Comparing constant term, d + 2 = 2c + d

c = 1 and d is arbitrary


Let f(x) = x

,x 1.x 1

Then for what value of is f(f(x)) = x?

(A) 2 (B) 2 (C) 1 (D) – 1(I I TJ EE 2001)

Example:

Solution :

xf x ,

x 1

x 1

ff x x 2x

x1 x 1

xx 1

xx

1x 1

2 21 x 1 x 0

This is an identity in x

= 1

+ 1 = 0 and 1 2 = 0


13

Example:

Solution :

I f f be the greatest integer function and g be the modulus

function, then find the value of (gof)53

(fog)53

f(x) = [x] and g(x) = |x|5 5

(gof) (fog)3 3

= 5 5

g ff g3 3

= 5 5

g f3 3

= 5g 2 f

3

= 2 – 1 = 5

| 2 |3

= 1


14

Let f(x) =1 | x | , x 1

[x], x 1

, where [.] denotethe

greatest integer function. Then find the value of f(f( 2.3))

Example:

Solution : f(x) = 1 | x | , x 1

[x], x 1

.

First find f( 2.3)

For x < –1, f(x) = 1 + |x|

f( 2.3) = 1 + | 2.3|

f( 2.3) = 3.3

Now f(f( 2.3)) = f(3.3)

For x –1, f(x) = [x]

f(f(2.3)) = f(3.3) = [3.3] = 3


15

I f g(x) = x2 + x 2 and 2gof(x) 4x 10x 4 , then find f(x)

Example:

Solution :2g(f(x)) 4x 10x 4 ……(i)

Also from g(x) = x2 + x 2

g(f(x)) = (f(x))2 + f(x) – 2 ……(ii)

Comparing (i) and (ii)

(f(x))2 + f(x) 2 = 4x2 10x + 4

f(x)2 + f(x) (4x2 – 10x + 6) = 0

f(x) = 21 1 4(4x 10x 6)

2

21 16x 40x 252


16

1 (4x 5)2

= 2x – 3 or 2 – 2x


17

Suppose that g(x) = 1 + x and f(g(x)) = 3 + 2 x + x, then find function f(x)

Example:

Solution :

g(x) = 1 + x and f(g(x)) = 3 + 2 x + x …….(i)

f(1 + x ) = 3 + 2 x + x

Put 1 + x = y x = (y 1)2 f(y) = 3 + 2(y 1) + (y 1)2

= 2 + y2

f(x) = 2 + x2


18

Example:

Solution :

I f f(x) = sin x + cos x, g(x) = x2 1, then g(f(x)) is invertible in the domain

(A) 0,2

(B) ,

4 4

(C) ,2 2

(D)[0,]

(I I TJ EE 2004)

f(x) = sin x + cos x, g(x) = x2 1

g(f(x)) = (sin x + cos x)2 1

= sin2x + cos2x + 2sin x cos x – 1 = sin 2x


19

x4 4

We know that sin is invertible when /2 /2

g(f(x)) is invertible in 2x2 2

Education

Composite functions