HOW TO FIND THE RANGE OF COMPOSITE FUNCTIONS

Question 1 :

Let A, B, C ⊆ N and a function f : A -> B be defined by f(x) = 2x + 1 and g : B -> C be defined by g(x) = x² . Find the range of f o g and g o f .

Solution :

f o g  =  f[g(x)]

=  f[x²]

Now we apply x² instead of x in f(x).

f o g  =  2 x² + 1

y = 2x² + 1

Range :

{y | y = 2x² + 1 and x ∊ N}

g o f  =  g[f(x)]

=  g[2x + 1]

Now we apply 2x + 1 instead of x in g(x).

g o f  =  (2x + 1)²

Range :

{y | y = (2x + 1)² and x ∊ N}

Question 2 :

Let f (x) = x² −1 . Find (i) f o f (ii) f o f o f

Solution :

(i) f o f  

  =  f[f(x)]

  =  f[x² −1]

Now we apply x² −1 instead of x in f(x).

  =  (x² −1)² - 1

  =  x⁴ - 2x² + 1 - 1

f o f  =  x⁴ - 2x² 

(ii) f o f o f

f o f  =  x⁴ - 2x² 

f o f o f  =  f [f o f]

  =  f[x⁴ - 2x²]

Now we apply x⁴ - 2x² instead of x in f(x).

  =  (x⁴ - 2x²)² - 1

Question 3 :

If f : R -> R and g : R -> R are defined by f(x) = x⁵ and g(x) = x⁴ then check if f, g are one-one and f o g is one-one?

Solution :

 f(x) = x⁵

For every positive and negative values of x, we get positive and negative values of y.

Every element in x is associated with different elements of y. Hence it is one to one function.

g(x) = x⁴ 

For every positive and negative values of x, we get only positive values of y.

Negative values of y is not associated with any elements of x. Hence it is not one to one function.

fog(x)  =  f[g(x)]

  =  f[x⁴]

now, we apply x⁴ instead of x in f(x)

f[x⁴]  =  (x⁵)⁴

fog(x)  =  x²⁰

fog is not one to one function.

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