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 :

f(x)  =  |x| + x

  =  x + x  (or)  -x + x

f(x)  =  2x  (or) f(x)  =  0

g(x)  =  |x| - x

  =  x - x  (or)  -x - x

g(x)  =  0  (or) g(x)  =  -2x

If f(x) = 2x and g(x)  =  0

gof =  g [f(x)]

  =  g [f(x)]

  =  g[2x]

gof  =  0

If f(x) = 0 and g(x) = -2x

gof =  g [f(x)]

  =  g [0]

gof  =  0

If f(x) = 2x and g(x)  =  0

fog =  f [g(x)]

  =  f [g(x)]

  =  f[0]

fog  =  0

If f(x) = 0 and g(x) = -2x

fog =  f [g(x)]

  =  f [-2x]

fog  =  0

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) = −16t2Graph 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.

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Problems on Finding Derivative of a Function

    Mar 29, 24 12:11 AM

    Problems on Finding Derivative of a Function

    Read More

  2. How to Solve Age Problems with Ratio

    Mar 28, 24 02:01 AM

    How to Solve Age Problems with Ratio

    Read More

  3. AP Calculus BC Integration of Rational Functions by Partical Fractions

    Mar 26, 24 11:25 PM

    AP Calculus BC Integration of Rational Functions by Partical Fractions (Part - 1)

    Read More