Questions on Functions with Solutions
Several questions on functions are presented and their detailed solutions discussed. The questions cover a wide range of concepts related to functions such as definition, domain, range, evaluation, composition and transformations of the graphs of functions.
Question 1Is the graph shown below that of a function?Solution to Question 1: Vertical line test: A vertical line at x = 0 for example cuts the graph at two points. The graph is not that of a function.
Question 2Does the equationrepresents a function y in terms of x? Solution to Question 2: Solve the above equation for y y^{ 2}= 1 - x y = + √(1 - x) or y = - √(1 - x) For one value of x we have two values of y and this is not a function.
Question 3Function f is defined byfind f(- 2). Solution to Question 3: Substitute x by -2 in the formula of the function and calculate f(-2) as follows f(-2) = - 2 (-2)^{ 2} + 6 (-2) - 3 f(-2) = -23
Question 4Function h is defined byfind h(x - 2). Solution to Question 4: Substitute x by x - 2 in the formula of function h h(x - 2) = 3 (x - 2)^{ 2} - 7 (x - 2) - 5 Expand and group like terms h(x - 2) = 3 ( x ^{ 2} - 4 x + 4 ) - 7 x + 14 - 5 = 3 x ^{ 2} - 19 x + 7 Question 5Functions f and g are defined byfind (f + g)(x)
Solution to Question 5:
Question 6Functions f and g are defined byfind (f + g)(x) and its domain. Solution to Question 6: (f + g)(x) is defined as follows (f + g)(x) = f(x) + g(x) = (1/x + 3x) + (-1/x + 6x - 4) Group like terms to obtain (f + g)(x) = 9 x - 4 The domain of function f + g is given by the intersection of the domains of f and g Domain of f + g is given by the interval (-∞ , 0) U (0 , + ∞)
Question 7Functions f and g are defined byfind (f / g)(x) and its domain. Solution to Question 7: (f / g)(x) is defined as follows (f / g)(x) = f(x) / g(x) = (x^{ 2} -2 x + 1) / [ (x - 1)(x + 3) ] Factor the numerator of f / g and simplify (f / g)(x) = f(x) / g(x) = (x - 1)^{ 2} / [ (x - 1)(x + 3) ] = (x - 1) / (x + 3) , x not equal to 1 The domain of f / g is the intersections of the domain of f and g excluding all values of x that make the numerator equal to zero. The domain of f / g is given by (-∞ , -3) U (-3 , 1) U (1 , + ∞)
Question 8Find the domain of the real valued function h defined bySolution to Question 8: For function h to be real valued, the expression under the square root must be positive or equal to 0. Hence the condition x - 2 ≥ 0 Solve the above inequality to obtain the domain in inequality form x ≥ 2 and interval form [2 , + ∞)
Question 9Find the domain ofSolution to Question 9: For a value of the variable x to be in the domain of function g given above, two conditions must be satisfied: The expression under the square root must not be negative - x^{ 2} + 9 ≥ 0 and the denominator of 1 / (x - 1) must not be zero x not equal to 1 or in interval form (-∞ , 1) ∪ (1 , + ∞) The solution to the inequality - x^{ 2} + 9 ≥ 0 is given by the interval [-3 , 3] Since x must satisfy both conditions, the domain of g is the intersection of the sets (-∞ , 1) ∪ (1 , + ∞) and [-3 , 3] [-3 , 1) ∪ (1 , +3]
Question 10Find the range of
Solution to Question 10:
Question 11Find the range ofSolution to Question 11: -x^{ 2} is either negative or equal to zero as x takes real values, hence -x^{ 2} <= 0 Add -10 to both sides of the above inequality to obtain -x^{ 2} - 10 <= -10 The expression on the left side is equal to f(x), hence f(x) <= -10 The above inequality gives the range of f as the interval (-∞ , -10]
Question 12Find the range ofSolution to Question 12: h(x) is a quadratic function, so let us first write it in vertex form using completing the square h(x) = x^{ 2} - 4 x + 9 = x^{ 2} - 4 x + 4 - 4 + 9 = (x - 2)^{ 2} + 5 (x - 2)^{ 2} is either positive or equal to zero as x takes real values, hence (x - 2)^{ 2} ≥ 0 Add 5 to both sides of the above inequality to obtain (x - 2)^{ 2} + 5 ≥ 5 The above inequality gives the range of h as the interval [5 , + ∞)
Question 13Functions g and h are given byFind the composite function (g _{o} h)(x). Solution to Question 13: The definition of the absolute value gives (g _{o} h)(x) = g(h(x)) = g(x^{ 2} + 1) = g(x^{ 2} + 1) = √(x^{ 2} + 1 - 1) = | x | So (g _{o} h)(x) = | x |
Question 14How is the graph of f(x - 2) compared to the graph of f(x)?Solution to Question 14: The graph of f(x - 2) is that of f(x) shifted 2 units to the right.
Question 15: How is the graph of h(x + 2) - 2 compared to the graph of h(x)?
Question 16Express the perimeter P of a square as a function of its area.Solution to Question 16: A square shape with side x has perimeter P given by P = 4 x and an area A given by A = x^{ 2} Solve the equation P = 4 x for x x = P / 4 and substitute into the formula for A to obtain A = (P / 4)^{ 2} = P ^{ 2} / 16 For any square shape the area A may be expressed as a function the perimeter P as follows A = P ^{ 2} / 16
Exercises
Answers to Above Exercises
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