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.
Questions with SolutionsQuestion 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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