COMPOSTION OF SQUARE ROOT FUNCTION

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

x  ≥ 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.