Questions on Functions with Solutions


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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 1: Is the graph shown below that of a function?

graph of question 1

Solution to Question 1:

  • Vertical line test: A vertcal line at x = 0 for example cuts the graph at two points. The graph is not that of a function.


Question 2: Does the equation

y 2 + x = 1


represents a function y in terms of x?

Solution to Question 2:

  • Solve the above equation for y

    y 2= 1 - x

    y = + SQRT(1 - x) or y = - SQRT(1 - x)

  • For one value of x we have two values of y and this is not a function.


Question 3: Function f is defined by

f(x) = - 2 x 2 + 6 x - 3


find 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 4: Function h is defined by

h(x) = 3 x 2 - 7 x - 5


find 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 5: Functions f and g are defined by

f(x) = - 7 x - 5 and g(x) = 10 x - 12


find (f + g)(x)

Solution to Question 5:

  • (f + g)(x) is defined as follows

    (f + g)(x) = f(x) + g(x) = (- 7 x - 5) + (10 x - 12)

  • Group like terms to obtain

    (f + g)(x) = 3 x - 17


Question 6: Functions f and g are defined by

f(x) = 1/x + 3x and g(x) = -1/x + 6x - 4


find (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 (-infinity , 0) U (0 , + infinity)


Question 7: Functions f and g are defined by

f(x) = x 2 -2 x + 1 and g(x) = (x - 1)(x + 3)


find (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

    (-infinity , -3) U (-3 , 1) U (1 , + infinity)


Question 8: Find the domain of the real valued function h defined by

h(x) = SQRT ( x - 2)

Solution 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 , + infinity)


Question 9: Find the domain of

g(x) = SQRT ( - x 2 + 9) + 1 / (x - 1)

Solution 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 denomirator of 1 / (x - 1) must not be zero

    x not equal to 1

    or in interval form

    (-infinity , 1) U (1 , + infinity)

  • 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
    (-infinity , 1) U (1 , + infinity) and [-3 , 3]


    [-3 , 1) U (1 , +3]


Question 10: Find the range of

f(x) = | x - 2 | + 3

Solution to Question 10:

  • | x - 2 | is an absolute value and is either positive or equal to zero as x takes real values, hence

    | x - 2 | >= 0

  • Add 3 to both sides of the above inequality to obtain

    | x - 2 | + 3 >= 3

  • The expression on the left side of the above inequality is equal to f(x), hence

    f(x) >= 3

  • The above inequality gives the range as the interval

    [3 , + infinity)


Question 11: Find the range of

f(x) = -x 2 - 10

Solution 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

    (-infinity , -10]


Question 12: Find the range of

h(x) = x 2 - 4 x + 9

Solution 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 , + infinity)


Question 13: Functions g and h are given by

g(x) = SQRT(x - 1) and h(x) = x 2 + 1


Find 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)

    = SQRT(x 2 + 1 - 1)

    = | x |

  • So

    (g o h)(x) = | x |


Question 14: How 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)?

Solution to Question 15:

  • The graph of h(x + 2) - 2 is that of h(x) shifted 2 units to the left and 2 unit downward.


Question 16: Express 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:

  1. Evaluate f(3) given that f(x) = | x - 6 | + x 2 - 1

  2. Find f(x + h) - f(x) given that f(x) = a x + b

  3. Find the domain of f(x) = SQRT(-x 2 - x + 2)

  4. Find the range of g(x) = - SQRT(- x + 2) - 6

  5. Find (f o g)(x) given that f(x) = SQRT(x) and g(x) = x 2 - 2x + 1

  6. How do you obtain the graph of - f(x - 2) + 5 from the graph of f(x)?

Answers to Above Exercises:

  1. f(3) = 11

  2. f(x + h) - f(x) = a h

  3. [-2 , 1]

  4. (- infinity , - 6]

  5. (f o g)(x) = | x - 1 |

  6. Shift the graph of f 2 units to the right then reflect it on the x axis, then shift it upward 5 units.


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Updated: 2 April 2013

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