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.
Examples with Solutions and Explanations on Operations on Functions
Example 1Function \( f \) is given by a table of values and function \( g \) is graphed below.
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 1a) \( (f + g)(2) = f(2) +g(2) \)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 2Let \( f(x) = 2x^2 + 3x + 1 \) 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 2a)\( (f + g)(0) = f(0) + g(0) = (2(0)^2 + 3(0) + 1) + (3(0) + 2) = 3\) \( (f - g)(-1) = f(-1) - g(-1) = (2(-1)^2 + 3(-1) + 1) - (3(-1) + 2) = 1\) \( (f \cdot g) (1) = f(1) \cdot g(1) = (2(1)^2 + 3(1) + 1) \cdot (3(1)+2) = 30\) \( \left(\dfrac{f}{g} \right)(2) = \dfrac{f(2)}{g(2)} = \dfrac{2(2)^2 + 3(2) + 1}{3(2)+2} = \dfrac{15}{8} \) \( \left(\dfrac{g}{f} \right)(-1) = \dfrac{g(-1)}{f(-1)} = \dfrac{3(-1) + 2}{2(-1)^2 + 3(-1) + 1} = \dfrac{-1}{0} = \text{undefined} \) 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 3Let \( f(x) = \sqrt{x-1} \) and \( g(x) = x - 5 \).Find the domains of \( f + g \), \( f - g \), \( f \cdot g \) and \( \dfrac{f}{g} \). Solution to Example 3We 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. 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 4Let \( f(x) = \sqrt{x+1} \) 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 4The 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. 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 5Let \( f(x) = \dfrac{1-x}{x+2} \) 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 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 FunctionsQuestion 1Functions \( f \) and \( g \) are given by table of values.
\( (f - g)(-1) \) , \( (g + f)(0) \) , \( (f \cdot g)(2) \) , \( (g \cdot f)(1) \) , \( (\dfrac{g}{f})(1) \) , \( (\dfrac{f}{g})(1) \) b) Find the domains of of \( f + g \), \( f - g \), \( f \cdot g \) , \( \dfrac{f}{g} \) and \( \dfrac{g}{f} \)
Question 2Let \( f(x) = \sqrt{-x-5} \) and \( g(x) = |x - 7| \).Evaluate \( (f + g)(0) \) , \( (f - g)(-6) \), \( (f \cdot g) (-9) \) and \( \left(\dfrac{f}{g} \right)(-30) \).
Question 3Let \( f(x) = 2x + 14 \) 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 4Let \( f(x) = \dfrac{x-2}{x+3} \) 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 1a)\( (f - g)(-1) = f(-1) - g(-1) = 10 - (-4) = 14\) \( (g + f)(0) = g(0) + f(0) = \) 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 4Let \( f(x) = \dfrac{x-2}{x+3} \) 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: \( (-\infty , - 3) \cup (-3 , 1) \cup (1 , 2) \cup (2 , +\infty) \) More References and linksOperations on Functions - Graphing CalculatorComposition of Functions functions Questions on Composite Functions with Solutions. Maths Problems, Questions and Online Self Tests |