PROBLEMS ON COMPOSITE FUNCTIONS

Problem 1 :

If f, g : R → R are defined by f(x) = |x| + x and g(x) = |x| − x, find g ◦ f and f ◦ g.

Solution :

For all values of x, fog and gof is 0.

Problem 2 :

If f, g, h are real valued functions defined on R, then prove that (f + g) ◦ h = f◦h + g ◦ h. What can you say about f ◦ (g + h) ? Justify your answer.

Solution :

 (f + g) ◦ h  =   (f + g) [h(x)]

  =  f [h(x)] + g [h(x)]

  =  f◦h + g ◦ h  ----(1)

f ◦ (g + h)  =  f [(g + h)(x)]

  =  f[g(x) + h(x)]

  =  f[g(x)] + f[h(x)]

  =  f ◦ g + f ◦ h  ----(2)

It is a function.

Problem 3 :

If f : R → R is defined by f(x) = 3x − 5, prove that f is a bijection and find its inverse.

Solution :

f(x)  =  3x − 5

f : A → B is said to be bijective if it is both one-to-one and onto.

For all x ∊ A, we get different values of y ∊ B, so it is one to one function.

For all y ∊ B, there is a preimage in A. So it is onto function.

Hence it is bijective function.

Inverse function :

Let y = 3x - 5

3x  =  y + 5

x  =  (y + 5)/3

f^-1(x)  =  (x + 5)/3

Problem 4 :

The weight of the muscles of a man is a function of his body weight x and can be expressed as W(x) = 0.35x. Determine the domain of this function.

Solution :

Given that :

W(x) = 0.35x

Here "x" represents weight of the body, it will not be negative. Hence its domain will be > 0.

Problem 5 :

The distance of an object falling is a function of time t and can be expressed as s(t) = −16t². Graph the function and determine if it is one-to-one.

Solution :

From the above graph, we come to know that for different values of t  ∊ A, we get different value of s(t) ∊ B. Hence it is one to one function.

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