# Composition of Functions

Let A B and C be three non-empty sets. Let f →A : B and g →B : C be two mappings (or functions), then gof → A : C. This function is called the product or composite of f and g given by

go*f*(x) = g{*f*(x)} ∀ x $\in$ A

gof exists iff the range of f is a subset of domain of g. Similarly, fog exists if range of g is a subset of domain of f .

**Properties of Composition of Function**

(i) The composition of functions is not commutative.

i.e.,** fog $\neq$ gof**

(ii) The composition of functions is associative.

i.e., *f*o(goh) = (*f*og)oh

(iii) The composition of any function with the identity function is the function itself.

i.e., If *f* : A → B, then *f*o$I_A$ = $I_B$o*f* = *f*

**Sample Problem 1**

If f (x) = 5x and g(x) = x+3, then find (f∘g)(x) if x = 2.

Answer: Given, f(x) = 5x

g(x) = x+ 3

Therefore, the composition of f from g will be;

(f∘g)(x) = f(g(x)) = f(x+3) = 5(x+3)

Now putting the value of x = 2

f(g(2)) = 5(2+3) = 5(5) = 25

**Sample Problem 2**

If f(x) = x +1 and g(x) = -2x^{2}, then find (g∘f)(x) for x = 1.

Answer: Given,

f(x) = x+1

g(x) = -2x^{2}

To find: g(f(x))

g(f(x)) = g(x+1) = -2(x+1)^{2}

Now put x =1 to get;

g(f(1)) = -2(1+1)^{2}

= -2(2)^{2}

= -8

**Sample Problem 3**

If f →R : R, g →R : R and h →R : R are such that
f(x)= x$^2$, g(x) = tan x and h(x) = log x, then the
value of (ho(gof))(x) , if x =
$\frac{\pi}{4}$

(a) 0

(b) 1

(c) –1

(d) p

Answer: (a)

(ho(gof))(x) = ho{g(x$^2$)}= ho(tan x$^2$) = log (tan x$^2$)

at x = $\frac{\pi}{4}$

(ho(gof))(x) = log $\left(tan\left(\frac{\pi}{4}\right)^2\right)$

(ho(gof))(x) = log $\left(tan \frac{\pi}{4}\right)$ = log 1 = 0

**Sample Problem 4**

If there are three functions, such as f(x) = x, g(x) = x^{2} and
h(x) = 4x. Then find the composition of these functions such as [f ∘ (g
∘ h)] (x) for x = -2.

Solution: Given,

f(x) = x, g(x) = x^{2} and h(x) = 4x

To find: [f ∘ (g ∘ h)] (x)

[f ∘ (g ∘ h)] (x) = f ∘ (g(h(x)))

= f ∘ g(4x)

= f((4x)^{2})

= f(16x^{2})

= 16x^{2}

If x = -2, then;

[f ∘ (g ∘ h)] (-2) = 16(-2)^{2} = 64

**Sample Problem 5**

Suppose *f*(x) gives miles that can be driven in *x *hours and g(y) gives the gallons of gas used in driving y miles. Which of these expressions is meaningful: *f*(g(*y*)) or g(*f*(x))?

**Answer:**

The function y = f(x)

is a function whose output is the number of miles driven corresponding to the number of hours driven.

The function g(y)

is a function whose output is the number of gallons used corresponding to the number of miles driven. This means:

The expression g(y)

takes miles as the input and a number of gallons as the output. The function f(x) requires a number of hours as the input. Trying to input a number of gallons does not make sense. The expression

is meaningless.

The expression f(x)

takes hours as input and a number of miles driven as the output. The function requires a number of miles as the input. Using f(x) (miles driven) as an input value for g(y), where gallons of gas depends on miles driven, does make sense. The expression g(*f*(x)) makes sense, and will yield the number of gallons of gas used, driving a certain number of miles, f(x) in *x *hours.

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