# COMPOSTION OF SQUARE ROOT FUNCTION

The composition of a function g with a function f is :

h(x) = g(f (x))

The domain of h is the set of all x-values such that x is in the domain of f and f (x) is in the domain of g.

Example 1 :

Let f(x) = x2 - 1 and let g(x) = √x. Find the domain of the composition functions

(i)  g∘f

(ii)  f∘g

Solution :

Domain of f(x) :

f(x) = x2 - 1

All real values, so domain is (-∞, ∞)

Domain of g(x) :

g(x) = √x

All positive values, so domain is [0, ∞)

(i)  g∘f

g∘f  = g( f(x) )

= g(x2 - 1)

Here instead of x, we have  x2 - 1. So, in the function g(x) we have to apply x2 - 1.

g∘f  = √(x2 - 1).

Domain of g∘f :

√(x2 - 1)  ≥ 0

(x2 - 1)  ≥ 0

(x + 1) (x - 1)  ≥ 0

≥ 1 and  x ≥ -1

Comparing domain of f(x) and this domain the intersection part is (-∞, -1] and [1, ∞).

So the required domain for  g∘f is (-∞, -1] U [1, ∞).

(ii)  f∘g

f∘g  = f( g(x) )

= f(√x)

Here we see √x instead of x, so we will apply x as √x in the function f(x).

f∘g = (√x)2 - 1

x - 1

Domain of f∘g :

Domain of x - 1 is all real values.

Comparing domain of g(x) and this domain the intersection part is [0, ∞).

So, the domain f∘g is [0, ∞).

Example 2 :

Let

f(x) = 1/(x2 - 1) and g(x) = √(x - 2)

Find the domain of f(g(x)).

Solution :

Domain of g(x) is real values ≥ 2

Domain of 1/(x - 3) is all real values except 3.

So, the required domain is f(g(x)) is [2, 3) U (3, ∞).

Example 3 :

Let

f(x) = √(x + 2) and g(x) = x2

Find the domain of f(g(x)).

Solution :

f(g(x)) = f(x2)

Here instead of x, we have x2. So in the function f(x), we have to apply the value x2

√(x2 + 2)

Even though we give negative values of x, f(g(x)) will give us positive result.

So, the domain of f(g(x)) is all real values.

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