# DOMAIN OF COMPOSITION OF FUNCTIONS

To find domain of composition of function, we have to consider the following.

i) Find composition of functions.

ii) If the composition function is polynomial function, that will be defined for all real values. So, domain will be  (-∞, ∞).

iii) If the composition function is square root function, then after applying the value of x, we should not receive negative values inside the square root. On this condition, we have to fix the domain.

iv) If we have rational function, the denominator should never become 0. The values which are making the denominators 0, we have to exclude those values from domain.

For each of the following problems :

(a)  Find f ∘ g and its domain.

(b) Find g ∘ f and its domain.

Problem 1 :

f(x) = x2 + 3x; g(x) = 2x - 7

Solution :

(a) To find f ∘ g and its domain :

f ∘ g = f[g(x)]

= f[2x - 7]

= (2x - 7)2 + 3(2x - 7)

= (2x)2 + 72 - 2(2x)(7) + 6x - 21

= 4x2 + 49 - 28x + 6x - 21

(f ∘ g)(x) = 4x2 - 22x + 28

Domain is all real values of x.

So, domain of f ∘ g is (-∞, ∞).

(b) To find g ∘ f and its domain :

g ∘ f = g[f(x)]

= g[x2 + 3x]

= 2(x2 + 3x) - 7

(g ∘ f)(x) = 2x2 + 6x - 7

Domain is all real values of x.

So, domain of g ∘ f is (-∞, ∞).

Problem 2 :

f(x) = 6x + 2; g(x) = 7 - x2

Solution :

(a) To find f ∘ g and its domain :

f ∘ g = f[g(x)]

= f[7 - x2]

= 6(7 - x2) + 2

= 42 - 6x2 + 2

(f ∘ g)(x) = 44 - 6x2

So, domain of f ∘ g is (-∞, ∞).

(b) To find g ∘ f and its domain :

g ∘ f = g[f(x)]

= g[6x + 2]

= 7 - (6x + 2)2

= 7 - [(6x)2 + 22 + 2(6x)(2)]

= 7 - [36x2 + 4  + 24x]

= 7 - 36x2 - 4 - 24x

(g ∘ f)(x) = -36x2 - 24x  + 3

So, domain of g ∘ f is (-∞, ∞).

Problem 3 :

Solution :

(a) To find f ∘ g and its domain :

f ∘ g = f[g(x)]

So, domain of f ∘ g is all real numbers expect 4.

(b) To find g ∘ f and its domain :

g ∘ f = g[f(x)]

= g(x2)

√(x- 4) ≥ 0

(x- 4) ≥ 0

(x + 2)(x - 2) ≥ 0

By drawing number line, we will get the solution more clearly.

x = -3  (-∞, -2)

Appling the above value in (x + 2) (x - 2) ≥ 0, we get

(-3+2)(-3-2) ≥ 0

True

x = 0  (-2, 2)

Appling the above value in (x + 2) (x - 2) ≥ 0, we get

(0+2)(0-2) ≥ 0

False

x = 4  (2, ∞)

Appling the above value in (x + 2) (x - 2) ≥ 0, we get

(4+2)(4-2) ≥ 0

True

So, domain of g ∘ f is (-∞, -2) ∪ (2, ∞).

Problem 4 :

Solution :

(a) To find f ∘ g and its domain :

f ∘ g = f[g(x)]

= f(x)2

Domain is all real values. Because no values will make the denominator as zero. Then the entire function will not become undefined for any values.

(b) To find g ∘ f and its domain :

g ∘ f = g[f(x)]

So, domain of g ∘ f is all real numbers except -5.

Problem 5 :

f(x) = √(x + 7); g(x) = -5 - x

Solution :

(a) To find f ∘ g and its domain :

f ∘ g = f[g(x)]

= f[-5 - x]

= √(-5 - x + 7)

(f ∘ g)(x) = √(2 - x)

So, domain of f ∘ g is  (-∞, 2]

(b) To find g ∘ f and its domain :

g ∘ f = g[f(x)]

= g[√(x + 7)]

(g ∘ f)(x) = -5 - √(x + 7)

So, domain of g ∘ f is  [-7, ∞).

Problem 6 :

f(x) = √(3 - x); g(x) = 9 - 2x

Solution :

(a) To find f ∘ g and its domain :

f ∘ g = f[g(x)]

= f[9 - 2x]

= √(3 - (9 - 2x))

= √(3 - 9 + 2x)

(f ∘ g)(x) = √(2x - 6)

So, domain of f ∘ g is [3, ∞).

(b) To find g ∘ f and its domain :

g ∘ f = g[f(x)]

= g[√(3 - x)]

= 9 - 2(√(3 - x))

(g ∘ f)(x) = 9 - 2√3 + 2x

So, domain of g ∘ f is (-∞, ∞).

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