# EXPRESS THE FUNCTION AS A COMPOSITION OF THREE FUNCTIONS

## About "Express the Function as a Composition of Three Functions"

Express the Function as a Composition of Three Functions :

Here we are going to see, how to express the function as a composition of three functions.

Question 1 :

Consider the functions f (x), g(x), h(x) as given below. Show that (f o g) o h = f o (g o h) in each case.

(i) f(x) = x −1, g(x) = 3x +1 and h(x) = x2

Solution :

(f o g) o h

fog(x)  =  f[g(x)]

=  f[3x + 1]

=  3x + 1 - 1

fog(x)  =  3x

(f o g) o h  =  (f o g) [h(x)]

=  (f o g) [x2]

(f o g) o h  =  3x2     --------(1)

f o (g o h)

(g o h)  =  g[h(x)]

=  g[x2]

=  3x2 +1

f o (g o h)  =  f [goh]

=  f[3x2 +1]

=  3x2 + 1 - 1

f o (g o h)  =  3x-------(2)

(1)  =  (2)

(f o g) o h  =  f o (g o h)

(ii) f (x) = x2, g(x) = 2x and h(x) = x + 4

Solution :

(f o g) o h

fog(x)  =  f[g(x)]

=  f[2x]

=  (2x)2

fog(x)  =  4x2

(f o g) o h  =  (f o g) [h(x)]

=  (f o g) [x + 4]

(f o g) o h  =  4(x+4)2

=  4(x2 + 8x + 16)

=  4x2 + 32x + 64     --------(1)

f o (g o h)

(g o h)  =  g[h(x)]

=  g[x + 4]

=  2(x + 4)

(g o h)  =  2x + 8

f o (g o h)  =  f [goh]

=  f[2x + 8]

f o (g o h)  =  (2x + 8)2

=  (2x)2 + 2(2x)(8) + 82

=  4x2 + 32 x + 64-------(2)

(1)  =  (2)

(f o g) o h  =  f o (g o h)

(iii) f (x) = x −4, g(x) = x2 and h(x) = 3x −5

Solution :

(f o g) o h

fog(x)  =  f[g(x)]

=  f[x2]

=  x2 −4

fog(x)  =  x2 −4

(f o g) o h  =  (f o g) [h(x)]

=  (f o g) [3x −5]

(f o g) o h  =  (3x - 5)2 - 4

=  (3x)2 - 2 (3x)(5) + 52 - 4

=  9x2 - 30x + 25 - 4

=  9x2 - 30x + 21   --------(1)

f o (g o h)

(g o h)  =  g[h(x)]

=  g[3x −5]

=  (3x −5)2

=  (3x)2 - 2 (3x)(5) + 52

(g o h)  =  9x2 - 30x + 25

f o (g o h)  =  f [goh]

=  f[9x2 - 30x + 25]

f o (g o h)  =  9x2 - 30x + 25 - 4

=  9x2 - 30x + 21   --------(2)

(1)  =  (2)

Hence proved.

Question 2 :

Let f = {(−1, 3),(0,−1),(2,−9)} be a linear function from Z into Z . Find f (x).

Solution :

Let the linear function be "y = ax + b"

 if x = -1, then y = 33  =  a(-1) + b3  =  -a + b -a + b  =  3   -------(1) if x = 0, then y = -1-1  =  a(0) + b-1  =  b b = -1

By applying the value of b in (1)

-a + (-1)  =  3

-a  =  3 + 1

-a  =  4

a  = -4

By applying the value of "a" and "b", we get

y = -4x - 1

Hence the required linear equation is -4x - 1.

Question 3 :

In electrical circuit theory, a circuit C(t) is called a linear circuit if it satisfies the superposition principle given by C(at1 + bt2) = aC(t1) + bC(t2), where a,b are constants. Show that the circuit C(t) = 3t is linear.

Solution :

Let's take two points  t1 and t2 from domain of C(t).

now,  c(at1 + bt2)  =  3(at1 + bt2)

c(at1) =  3at1

ac(t1) =  3at1

bc(t2) =  3bt2

3(at1 bt2)  =  3at+ 3at2

(or) C(at1 + bt2) = aC(t1) + bC(t2),

Hence c(t) is linear.

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