Example 1. Let the domain of a function be this set of values:
A = {0, 1, 2, −2}
and let the variable x assume each value. Let the rule that relates the value of y to the value of x be the following:
y = x² + 1.
a) Write the set of ordered pairs (x, y) which "represents" this function.
Answer. {(0, 1), (1, 2), (2, 5), (−2, 5)}
That is, when x = 0, then y = 0² + 1 = 1.
When x = 1, then y = 1² + 1 = 2. And so on.
b) Write the set B which is the range of the function.
Answer. B = {1, 2, 5, 5}. The values in the range are simply those values of y that correspond to each value of x.
Notice that to each value of x in the domain there corresponds one -- and only one -- value of the function. Even though the value 5 is repeated, it is still one and only one value.
Example 2. Here is a relationship in which y is not a function of x:
y² = x
When x = 4, for example -- y² = 4 -- then
y = 2 or −2. To each value of x, there is more than one value of y.
Problem 1. Let y be a function of x as follows:
y = 3x²
a) Which is the independent variable and which the dependent variable?
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x is the independent variable, y is the dependent.
b) The domain of a function are the values of the independent variable,
b) which are the values of x.
c) What is the natural domain of that function?
Since there is no natural restriction on the values of x, the natural domain of that function is any real number. x could take any value on the x-axis.
d) The range of a function are the values of the dependent variable,
d) which are the values of y.
e) What is the range of that function? (Consider that the values of x² are
e) never negative.)
y ≥ 0
(If you are not viewing this page with Internet Explorer 6, then your browser may not be able to display the symbol ≥, "is greater than or equal to;" or ≤, "is less than or equal to.")
f) Write any three values of that function as members of an ordered pair.
For example, (1, 3), (2, 12), (3, 27)
A function of a function
Again, let us consider these functions:
f(x) = x² + 1 and g(x) = 5x.
And now consider this function,
f(g(x)).
"f of g of x"
f has g as its argument:
f(g(x)) = f(5x) = (5x)² + 1 = 25x² + 1.
Again, f squares its argument and adds 1.
Now let's look at
g(f(x)).
g will operate on f. What does g do to its argument? It simply multiplies the argument by 5. Therefore,
g(f(x)) = g(x² + 1) = 5(x² + 1) = 5x² + 5.
The parentheses in g(x² + 1) are the parentheses of the functional notation. The parentheses of 5(x² + 1), however, are the grouping parentheses, which here indicate multiplication by 5.
Problem 2. Read each symbol.
Problem 7. Function of a function.
Let f(x) = x² and
g(x) = x + h. Write the function
a) f(g(x))
= f(x + h) = (x + h)² = x² + 2xh + h²
b) g(f(x))
= g(x²) = x² + h
Problem 8. Let f(x) = x − 3, and g(x) = 3 − x. Write the functions
f(g(x)) and g(f(x)).
f(g(x)) = f(3 − x) = (3 − x) − 3 = −x
g(f(x)) = g(x − 3) = 3 − (x − 3) = 3 − x + 3 = 6 − x
Problem 9. Let f(x) = x5 and g(x) = x1/5. Write the functions
f(g(x)) and g(f(x)).
f(g(x)) =
f(x1/5) = (x1/5)5 = x1 = x
g(f(x)) = g(x5) = (x5)1/5 = x
In Line 2) we removed the parentheses in the numerator, and multiplied by the reciprocal of the denominator. (Lesson 22 of Algebra.)