Operations on Functions

The operations on functions are presented with examples including detailed solutions and explanations. Solve and understand the examples before attempting the questions which also have detailed solutions.

Operations on Functions Definition and Domain

Functions with real output may be combined using addition, subtraction, mulitplication and division as we do with real numbers.
Let \( f \) and \( g \) be two functions with real outputs. The functions \( f + g \), \( f - g \), \( f \cdot g \) and \( \dfrac{f}{g} \) are defined as follows:
1) \( (f + g)(x) = f(x) + g(x) \) ; addition of two functions , sum of two functions
2) \( (f - g)(x) = f(x) - g(x) \) ; subtraction of two functions , difference of two functions
3) \( (f \cdot g)(x) = f(x) \cdot g(x) \) ; multiplication of two functions , product of two functions
4) \( \left(\dfrac{f}{g} \right)(x) = \dfrac{f(x)}{g(x)} \) ; division of two functions , quotient of two functions

The domain of \( f + g \), \( f - g \) and \( f \cdot g \) is the set \( D \) of all values of \( x \) that are common to the domain of \( f \) and the domain of \( g \). If \( D_f \) is the domain of \( f \) and \(D_g \) is the domain of \(g \) then \( D \) is the intersection of \( D_f \) and \(D_g\).
\[ D = D_f \cap D_g \]
The domain of \( \dfrac{f}{g} \) is the set \(D\) such that \( g(x) \ne 0 \).
Note: Two functions may also be combined by composition of functions which is studied somewhere else in this wenbsite.

Examples with Solutions and Explanations on Operations on Functions

Example 1

Function \( f \) is given by a table of values and function \( g \) is graphed below.

\( x \)	\( f(x) \)
-2	-4
-1	10
0	7
1	8
2	11

Evaluate, if possible, the following:
a) \( (f + g)(2) \) , b) \( (g - f)(0) \) , c) \( (f \cdot g)(-2) \) , d) \( (g \cdot f)(1) \) , e) \( (\dfrac{f}{g})(1) \) , f) \( (\dfrac{f}{g})(-1) \)

Solution to Example 1

a) \( (f + g)(2) = f(2) +g(2) \)
Use the table of \( f \) to read the values of \( f(2) \) and the graph of g to read \( g(2) \) to the nearest integer.
\( f(2) = 11 \), \( g(2) = -3 \)
Substitute and evaluate
\( (f + g)(2) = f(2) + g(2) = 11 - 3 = 8\)
b) \( (g - f)(0) = g(0) - f(0) = - 3 - 7 = - 10\)
c) \( (f \cdot g)(-2) = f(-2) \cdot g(-2) = (-4)(5) = -20\)
d) \( (g \cdot f)(1) = g(1) \cdot f(1) = (-4)(8) = -32\)

e) \( \left(\dfrac{f}{g} \right)(1) = \dfrac{f(1)}{g(1)} = \dfrac{8}{-4} = - 2\)

f) \( \left(\dfrac{f}{g} \right)(-1) = \dfrac{f(-1)}{g(-1)} = \dfrac{10}{0} = \text{undefined} \) , division by zero not allowed.

Example 2

Let \( f(x) = 2x^2 + 3x + 1 \) and \( g(x) = 3x + 2 \).
a) Evaluate \( (f + g)(0) \) , \( (f - g)(-1) \), \( (f \cdot g) (1) \) , \( \left(\dfrac{f}{g} \right)(2) \) and \( \left(\dfrac{g}{f} \right)(-1) \)
b) Find the functions \( (f + g)(x) \), \( (f - g)(x) \), \( (f \cdot g)(x) \) and \( \left( \dfrac{f}{g} \right)(x) \).

Solution to Example 2

a)
\( (f + g)(0) = f(0) + g(0) = (2(0)^2 + 3(0) + 1) + (3(0) + 2) = 3\)
\( (f - g)(-1) = f(-1) - g(-1) = (2(-1)^2 + 3(-1) + 1) - (3(-1) + 2) = 1\)
\( (f \cdot g) (1) = f(1) \cdot g(1) = (2(1)^2 + 3(1) + 1) \cdot (3(1)+2) = 30\)

\( \left(\dfrac{f}{g} \right)(2) = \dfrac{f(2)}{g(2)} = \dfrac{2(2)^2 + 3(2) + 1}{3(2)+2} = \dfrac{15}{8} \)

\( \left(\dfrac{g}{f} \right)(-1) = \dfrac{g(-1)}{f(-1)} = \dfrac{3(-1) + 2}{2(-1)^2 + 3(-1) + 1} = \dfrac{-1}{0} = \text{undefined} \)
b)
\( (f + g)(x) = f(x) + g(x) = (2x^2 + 3x + 1 ) + (3x + 2) = 2x^2 + 6x + 3 \)
\( (f - g)(x) = f(x)- g(x) = (2x^2 + 3x + 1 ) - (3x + 2) = 2x^2 - 1 \)
\( (f \cdot g)(x) = f(x) \cdot g(x) = (2x^2 + 3x + 1 )\cdot(3x + 2) = 6x^3+13x^2+9x+2 \)
\( \left( \dfrac{f}{g} \right)(x) = \dfrac{f(x)}{g(x)} = \dfrac{2x^2 + 3x + 1}{3x+2} \)

Example 3

Let \( f(x) = \sqrt{x-1} \) and \( g(x) = x - 5 \).
Find the domains of \( f + g \), \( f - g \), \( f \cdot g \) and \( \dfrac{f}{g} \).

Solution to Example 3

We first need to find the domains of functions \( f \) and \( g \).
The domain \( D_f \) of \( f \) is the set of values of x satisfying the inequality: \( x-1 \ge 0 \) (radicand must be positive for \( f(x) \) to be real).
Solve the above inequality to obtain \( x \ge 1 \) or in interval form \( D_f = [1 , +\infty) \)
The domain \(D_g\) of g is the set of all real numbers or in interval form : \( D_g = ( -\infty , +\infty ) \)
The intersection of \( D_f\) and \( D_g\) is interpreted graphically below.
domain of operations on functions, example 3

domain of operations on functions, example 3

and is given by
\( D_f \cap D_g = [1 , +\infty) \)
The domains of \( f + g \), \( f - g \), \( f \cdot g \) is given by \( D_f \cap D_g = [1 , +\infty) \).
To find the domain of \( \dfrac{f}{g} \) is given by \( D_f \cap D_g = [1 , +\infty) \) and we also need satisfy the condition \( g(x) \ne 0 \) (division by zero not allowed) which gives \( x \ne 5 \)
The domain of \( \dfrac{f}{g} \) is given in interval notation by
\( [1 , 5 ) \cup ( 5 , +\infty \)

Example 4

Let \( f(x) = \sqrt{x+1} \) and \( g(x) = \sqrt{9-x^2} \).
Find the domains of \( f + g \), \( f - g \), \( f \cdot g \) and \( \dfrac{f}{g} \).

Solution to Example 4

The domain \( D_f \) of \( f \) is the solution set to the inequality: \( x+1 \ge 0 \) (radicand must be positive)
The solution set of the above inequality given by the interval form \( D_f = [ - 1 , +\infty) \)
The domain \( D_g \) of \( g \) is the solution set to the inequality: \( 9-x^2 \ge 0 \) (radicand must be positive)
The solution set of the above inequality given by the interval form \( D_g = [ - 3 , + 3] \).
The intersection of \( D_f\) and \( D_g\) is shown below graphically.
domain of operations on functions, example 4

domain of operations on functions, example 4

and is given by
\( D_f \cap D_g = [- 1 , + 3] \)
The domains of \( f + g \), \( f - g \), \( f \cdot g \) is given by \( D_f \cap D_g = [- 1 , + 3] \).
To find the domain of \( \dfrac{f}{g} \), we need to satisfy the condition \( g(x) \ne 0 \) (division by zero not allowed) which gives \( x \ne - 3 \) and \( x \ne + 3 \)
The domain of \( \dfrac{f}{g} \) is given by the intervals
\( [ - 1 , 3 )\)

Example 5

Let \( f(x) = \dfrac{1-x}{x+2} \) and \( g(x) = \dfrac{x-1}{x+4} \).
Find \( \dfrac{f}{g} \) and \( \dfrac{g}{f} \) and their domains.

Solution to Example 5

\( \left(\dfrac{f}{g} \right)(x) = \dfrac{f(x)}{g(x)} = \dfrac{\dfrac{1-x}{x+2}}{\dfrac{x-1}{x+4}} = \left (\dfrac{1-x}{x+2} \right) \left( \dfrac{x+4}{x-1} \right ) = - \left (\dfrac {x-1}{x+2} \right) \left( \dfrac{x+4}{x-1} \right ) = - \dfrac{x+4}{x+2} \)

\( \left(\dfrac{g}{f} \right)(x) = \dfrac{g(x)}{f(x)} = \dfrac{1}{\dfrac{f(x)}{g(x)}} = - \dfrac{x+2}{x+4} \)
The domain \( D_f \) of function \( f \) is given by the interval : \( ( - \infty , -2) \cup (-2 , +\infty) \)
The domain \( D_g \) of function \( g \) is given by the interval : \( ( - \infty , -4) \cup (-4 , +\infty) \)
The intersection of \( D_f \) and \( D_g \) is given by
\( D_f \cap D_g = D_g \cap D_f = ( - \infty , -4) \cup (-4 , -2 ) \cup (-2 , +\infty) \)
and shown graphically below
domain of operations on functions, example 5

domain of operations on functions, example 5

The domain of \( \dfrac{f}{g} \) is given by the interval \( ( - \infty , -4) \cup (-4 , -2 ) \cup (-2 , +\infty) \) such that \( g(x) \ne 0 \) or \( x \ne 1 \)
which is give by the interval \( ( - \infty , -4) \cup (-4 , -2 ) \cup (-2 , 1) \cup (1 , +\infty) \)

The domain of \( \dfrac{g}{f} \) is given by the interval \( ( - \infty , -4) \cup (-4 , -2 ) \cup (-2 , +\infty) \) such that \( f(x) \ne 0 \) or \( x \ne 1 \)
which is give by the interval \( ( - \infty , -4) \cup (-4 , -2 ) \cup (-2 , 1) \cup (1 , +\infty) \)

Questions with Solutions and Explanations on Operations on Functions

Question 1

Functions \( f \) and \( g \) are given by table of values.

\( x \)	\( f(x) \)	\( x \)	\( g(x) \)
-2	-4	-2	3
-1	10	-1	-4
0	7	1	9
1	0	2	-2
2	11	3	0

a) Evaluate, if possible, the following:
\( (f - g)(-1) \) , \( (g + f)(0) \) , \( (f \cdot g)(2) \) , \( (g \cdot f)(1) \) , \( (\dfrac{g}{f})(1) \) , \( (\dfrac{f}{g})(1) \)
b) Find the domains of of \( f + g \), \( f - g \), \( f \cdot g \) , \( \dfrac{f}{g} \) and \( \dfrac{g}{f} \)

Question 2

Let \( f(x) = \sqrt{-x-5} \) and \( g(x) = |x - 7| \).
Evaluate \( (f + g)(0) \) , \( (f - g)(-6) \), \( (f \cdot g) (-9) \) and \( \left(\dfrac{f}{g} \right)(-30) \).

Question 3

Let \( f(x) = 2x + 14 \) and \( g(x) = x + 7 \).
Find \( (f + g)(x) \) , \( (f - g)(x) \), \( (f \cdot g) (x) \), \( \left(\dfrac{f}{g} \right)(x) \) , \( \left(\dfrac{g}{f} \right)(x) \) and their domains.

Question 4

Let \( f(x) = \dfrac{x-2}{x+3} \) and \( g(x) = \dfrac{x+5}{x - 1} \).
Find the domains of \( \dfrac{f}{g} \) and \( \dfrac{g}{f} \).

Solutions to the Above Questions

Solution to Question 1

a)
\( (f - g)(-1) = f(-1) - g(-1) = 10 - (-4) = 14\)
\( (g + f)(0) = g(0) + f(0) = \) undefined, because \( g(0) \) is undefined
\( (f \cdot g)(2) = f(2) \cdot g(2) = (11)(-2) = -22 \)
\( (g \cdot f)(1) = g(1) \cdot f(1) = (9)(0) = 0 \)
\( (\dfrac{g}{f})(1) = \dfrac{g(1)}{f(1)} = \dfrac{9}{0} = undefined \)
\( (\dfrac{f}{g})(1) = \dfrac{f(1)}{g(1)} = \dfrac{0}{9} = 0\)
b)
The domain of f is given by the set \( D_f = \{ -2 , -1 , 0 , 1 , 2 \} \) and the domain of g is given by the set \( D_g = \{ -2 , -1 , 1 , 2 , 3\} \)
The domain of \( f + g \), \( f - g \), \( f \cdot g \) is given by the intersection of \( D_f \) and \( D_g \) given by
\( D_f \cap D_g = \{ -2 , -1 , 1 , 2 \} \)
The domain of \( \dfrac{f}{g} \) is the set \( D_f \cap D_g \) such that \( g(x) \ne 0 \) or \( x \ne 3 \) but \( 3 \) is not in the set \( D_f \cap D_g \).
Hence the domain of \( \dfrac{f}{g} \) is given by \( \{ -2 , -1 , 1 , 2 \} \)
The domain of \( \dfrac{g}{f} \) is the set \( D_g \cap D_f \) such that \( f(x) \ne 0 \) or \( x \ne 1 \) but \( 1 \) is in the set \( D_f \cap D_g \) and has to be omitted.
Hence the domain of \( \dfrac{g}{f} \) is given by \( \{ -2 , -1 , 2 \} \)

Solution to Question 2

\( (f + g)(0) = f(0) + g(0) = \sqrt{-(0) -5} + |0 - 7| = \sqrt{-5} + |- 7|\)not real, because of the term \( \sqrt{-5}\)
\( (f - g)(-6) = \sqrt{-(-6)-5} + |(-6) - 7| = \sqrt{1} + |-13| = 1 + 13 = 14 \)
\( (f \cdot g) (-9) = \sqrt{-(-9)-5} \cdot |(-9) - 7| = \sqrt{4} \cdot |-16| = (2)(16) = 32 \)

\( \left(\dfrac{f}{g} \right)(-30) = \dfrac{\sqrt{-(-30)-5}}{|(-30) - 7|} = \dfrac{\sqrt{25}}{|-37|} = \dfrac{5}{37} \)

Solution to Question 3

\( (f + g)(x) = f(x) + g(x) = 2x + 14 + x + 7 = 3 x + 21 \)
\( (f - g)(x) = f(x) - g(x) = (2x + 14) - (x + 7) = x + 7 \)
\( (f \cdot g) (x) = f(x) \cdot g(x) = (2x + 14)(x + 7) = 2x^2+28x+98 \)
\( \left(\dfrac{f}{g} \right)(x) = \dfrac{f(x)}{g(x)} = \dfrac{2x + 14}{x + 7} = \dfrac{2(x + 7)}{x + 7} = 2 , for \; x \ne -7\)
\( \left(\dfrac{g}{f} \right)(x) = \dfrac{g(x)}{f(x)} = \dfrac{x + 7}{2x + 14} = \dfrac{x + 7}{2(x + 7)} = 1/2 , for \; x \ne -7\)

The domain \( D_f \) of function \( f \) is given by: \( D_f = (-\infty,+\infty)\)
The domain \( D_g \) of function \( g \) is given by: \( D_g = (-\infty,+\infty)\)
The intesection of \( D_f \) and \( D_g \) is given by \( D = D_f \cap D_g = (-\infty,+\infty)\)
The domain of \( f + g \) , \( f - g \) and \( f \cdot g \) is \( D = (-\infty,+\infty) \)
The domain of \( \left(\dfrac{f}{g} \right) \) is \( D = (-\infty,+\infty)\) such that \( g(x) \ne 0 \) and \( x \ne -7 \) which is given by \( (-\infty, - 7 ) \cup (-7 , +\infty)\)
The domain of \( \left(\dfrac{g}{f} \right) \) is \( D = (-\infty,+\infty)\) such that \( f(x) \ne 0 \) and \( x \ne -7 \) which is given by \( (-\infty, - 7 ) \cup (-7 , +\infty)\)

Solution to Question 4

Let \( f(x) = \dfrac{x-2}{x+3} \) and \( g(x) = \dfrac{x+5}{x - 1} \).
Find the domains of \( \dfrac{f}{g} \) and \( \dfrac{g}{f} \).
The domain \( D_f \) of \( f \) is given by: \( D_f = (-\infty , - 3) \cup (-3 , +\infty) \)
The domain \( D_g \) of \( g \) is given by: \( D_g = (-\infty , 1) \cup (1 , +\infty) \)
The intersection \( D_f \cap D_g \) is given by \( D_f \cap D_g = (-\infty , - 3) \cup (-3 , 1) \cup (1 , +\infty) \)
The domain of \( \dfrac{f}{g} \) is the set \( D_f \cap D_g \) such that \( g(x) \ne 0 \) or \( x \ne -5 \) which is given by: \( (-\infty , - 3) \cup (-3 , -5) \cup (-5 , 1) \cup (1 , +\infty) \)
The domain of \( \dfrac{g}{f} \) is the set \( D_f \cap D_g \) such that \( f(x) \ne 0 \) or \( x \ne 2 \) which is given by: \( (-\infty , - 3) \cup (-3 , 1) \cup (1 , 2) \cup (2 , +\infty) \)

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Operations on Functions