Key Ideas :
• The domain of a function is the set of values of the independent variable for which the function is defined. The range of a function depends on the equation of the function. The graph depends on the domain and range.
• The domain and range of a function can be determined from its graph, from a table of values, or from the function equation. They are usually easier to determine from a graph or a table of values.
Need to Know :
• All linear functions include all the real numbers in their domains. Linear functions of the form f(x) = mx + b where m ≠ 0, have range {y ∊ R}. Constant functions f(x) = b have range {b}.
• All quadratic functions have domain x ∊ R. The range of a quadratic function depends on the maximum or minimum value and the direction of opening.
• The domains of square root functions are restricted because the square root of a negative number is not a real number. The ranges are restricted because the square root sign refers to the positive square root. For example,
(i) The function f(x) = √x has domain = {x ∊ R | x ≥ 0} and range {y ∊ R | y ≥ 0}.
(ii) The function g(x) = √(x - 3) has domain = {x ∊ R | x ≥ 3} and range {y ∊ R | y ≥ 0}.
• When working with functions that model real-world situations, consider whether there are any restrictions on the variables. For example, negative values often have no meaning in a real context, so the domain or range must be restricted to nonnegative values.
Example 1 :
f(x) = 5x + 9
Solution :
This is a linear function, so x and y can be any value.
Domain = {x ∊ R}
Range = {y ∊ R}
Example 2 :
g(x) = -3(x + 1)^{2} + 6
Solution :
This is a quadratic equation in vertex form. The function has a maximum value at the vertex (-1, 6), x can be any value.
Domain = {x ∊ R}
Range = {y ∊ R | y ≤ 6}
Example 3 :
h(x) = √(2 - x)
Solution :
We cannot take the square root of a negative number, so (2 - x) must be positive or zero.
2 - x ≥ 0
x ≤ 2
Domain = {x ∊ R | x ≤ 2}
√(2 - x) means the positive square root, so y is never negative.
Range = {y ∊ R | y ≥ 0}
Example 4 :
p(x) = 1/(x - 2)
Solution :
The given function is a rational function.
To find the domain of a rational function, we have to find the value of x that makes the denominator zero.
In 1/(x - 2), if we substitute x = 2, the denominator becomes zero and it is undefined.
So, p(x) is defined for all real values of x except x = 2.
Domain of p(x) = R - {0}
To find range of the rational function above, find the inverse of p(x).
p(x) = 1/(x - 2)
y = 1/(x - 2)
Interchange the variables.
x = 1/(y - 2)
Solve for y in terms of x.
(y - 2)x = 1
y - 2 = 1/x
y = 1/x + 2
y = (1 + 2x)/x
y = (2x + 1)/x
p^{-1}(x) = (2x + 1)/x
Find the domain of p^{-1}(x).
In (2x + 1)/x ,if we substitute x = 0, the denominator becomes zero and it is undefined.
So, p^{-1}(x) is defined for all real values of x except x = 0.
Domain of p^{-1}(x) = R - {0}
Range of p(x) = Domain of p^{-1}(x)
Range of p(x) = R - {0}
Example 5 :
A gull landing on the guardrail causes a pebble to fall off the edge. The speed of the pebble as it falls to the ground is a function is v(d) = √(2gd) where d is the distance, in meters, the pebble has fallen, v(d) is the speed of the pebble, in meters per second (m/s) and g is the acceleration due to gravity—about 9.8 meters per second squared (m/s^{2}). Determine the domain and range of v(d), the pebble’s speed.
Solution :
d = 0 when the pebble begins to fall, and d = 346 when it lands.
So, the domain is 0 ≤ d ≤ 346.
The pebble starts with speed 0 m/s.
v(0) = v(d) = √(2 ⋅ 9.8 ⋅ 0) = 0
When the pebble lands, d = 346.
v(346) = √(2 ⋅ 9.8 ⋅ 346)
= √6781.6
≈ 82.4
The domain is
{d ∊ R | 0 ≤ d ≤ 346}
and the range is
{v(d) ∊ R | 0 ≤ v(d) ≤ 82.4}
Example 6 :
Vitaly and Sherry have 24 m of fencing to enclose a rectangular garden at the back of their house.
a) Express the area of the garden as a function of its width.
b) Determine the domain and range of the area function.
Solution :
They need fencing on only three sides of the garden because the house forms the last side.
Let the width of the garden be x m.
Then the length is (24 - 2x).
Let A(x) be the area of the garden.
A(x) = x(24 - 2x)
= 24x - 2x^{2}
= -2x^{2} + 24x
= -2(x^{2} - 12x)
= -2[x^{2} - 2(x)(6) + 6^{2} - 6^{2}]
= -2[(x - 6)^{2} - 36]
A(x) = -2(x - 6)^{2} - 72
The smallest the width can approach is 0 m. The largest the width can approach is 12 m.
Domain = {x ∊ R | 0 < x < 12}
A(x) = -2(x - 6)^{2} - 72 is a quadratic equation in vertex form and it has a maximum value at the vertex (6, 72).
Range = {A(x) ∊ R | 0 < A(x) ≤ 72}
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