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?

__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 = + √(1 - x) or y = - √(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) = √ ( 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) = √ ( - 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) = √(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)

= √(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:__

Evaluate f(3) given that f(x) = | x - 6 | + x^{ 2} - 1

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

Find the domain of f(x) = √(-x^{ 2} - x + 2)

Find the range of g(x) = - √(- x + 2) - 6

Find (f o g)(x) given that f(x) = √(x) and g(x) = x^{ 2} - 2x + 1

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

__Answers to Above Exercises:__

f(3) = 11

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

[-2 , 1]

(- infinity , - 6]

(f _{o} g)(x) = | x - 1 |

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

More on calculus

questions with answers, tutorials and problems and Questions and Answers on Functions.