Domain of absolute value function :
Domain is set of all possible values of inputs. For absolute value function, all real values can be considered as possible inputs.
Domain of absolute value function is (-∞, ∞)
Range of absolute value function :
Set of possible outputs is range. The range can be found easily by finding direction of opening and vertex the absolute value function.
Vertex form of absolute value function :
y = a|x - h| + k
Here (h, k) is vertex and sign of a will decide the direction of opening.
For example,
y = |t - 9| Domain (-∞, ∞) y = |t - 9| a = 1 opens up Vertex is (9, 0) Range (0, ∞) |
y = -|9 - t| Domain (-∞, ∞) y = -|t - 9| a = -1 opens down Vertex is (9, 0) Range (-∞, 0) |
Find the domain and range of each of the following functions. Express answers in interval notation.
Problem 1 :
f(t) = | t |
Solution:
f(t) = | t |
Domain :
A set of all defined value of t is known as domain.
D(t) = t ∈ R
Domain: (-∞, ∞)
Range :
y = | t |
Vertex is (0, 0)
The shape V opens up, so the range is [0, ∞)
Problem 2 :
g(t) = | 9 - t |
Solution:
Domain:
A set of all defined value of t is known as domain.
D(t) = t ∈ R
Domain: (-∞, ∞)
Range:
y = |9 - t|
y = |t - 9|
By observing the vertex of the absolute value function, vertex is at (9, 0). The curve will open up.
Range: [0, ∞)
Problem 3 :
h(t) = 9 - |t|
Solution:
Domain:
A set of all defined value of t is known as domain.
D(t) = t ∈ R
Domain: (-∞, ∞)
Range:
h(t) = 9 - | t |
Vertex form of the function h(t),
h(t) = -|t| + 9
Vertex is at (0, 9). The curve will open down. So, the range is
(-∞, 9]
Problem 4 :
f(x) = | x |
Solution:
f(x) = | x |
Domain:
A set of all defined value of x is known as domain.
D(x) = x ∈ R
Domain: (-∞, ∞)
Range:
y = | x |
Vertex is at (, 0). The absolute value function opens up. So, the range is [0, ∞)
Problem 5 :
g(x) = | x | + 1
Solution:
Domain:
A set of all defined value of x is known as domain.
D(x) = x ∈ R
Domain: (-∞, ∞)
Range:
g(x) = |x| + 1
Vertex is at (0, 1) and the curve opens up. So, the range [1, ∞)
Problem 6 :
h(x) = | x + 1|
Solution:
Domain:
A set of all defined value of x is known as domain.
D(x) = x ∈ R
Domain: (-∞, ∞)
Range:
h(x) = | x + 1|
Vertex is at (-1, 0), the curve opens. So, the range is at [0, ∞)
Problem 7 :
f(x) = | x^{2} + 3 |
Solution:
Domain:
A set of all defined value of x is known as domain.
D(x) = x ∈ R
Domain: (-∞, ∞)
Range:
f(x) = | x^{2} + 3 |
Vertex is at (0, 3). The curve will open up. So, the range
Range: [3, ∞)
Problem 8 :
g(x) = | x^{2} + 3 | - 4
Solution:
Domain:
A set of all defined value of x is known as domain.
D(x) = x ∈ R
Domain: (-∞, ∞)
Range:
g(x) = | x^{2} + 3 | - 4
Vertex is at (0, 3). The the curve is moved 4 units down. So, 3 - 4 = -1. Then,
range is [-1, ∞)
Problem 9 :
h(x) = 2| x^{2} + 3 | + 5
Solution:
Domain:
A set of all defined value of x is known as domain.
D(x) = x ∈ R
Domain: (-∞, ∞)
Range:
Vertex is at (0, 3). The the curve is moved 5 units up. So, 3 + 5 = 8. Curve opens up. Then,
range is (8, ∞)
Problem 10 :
f(t) = | t^{2} + 6 |
Solution:
Domain:
A set of all defined value of x is known as domain.
D(t) = t ∈ R
Domain: (-∞, ∞)
Range:
Vertex is at (0, 6). The curve opens up. Then,
range is (6, ∞)
Problem 11 :
g(t) = | t^{2} + 6 | + 7
Solution:
Domain:
A set of all defined value of x is known as domain.
D(t) = t ∈ R
Domain: (-∞, ∞)
Range:
Vertex is at (0, 6). The curve opens up. From 6, we have to move the curve 7 units up.
6 + 7 ==> 13
Then,
range is (13, ∞)
Problem 12 :
h(t) = -1/3 |t^{2} + 6 | - 8
Solution:
Domain:
A set of all defined value of x is known as domain.
D(t) = t ∈ R
Domain: (-∞, ∞)
Range:
Vertex is at (0, 6). The curve opens up. From 6, we have to move the curve 8 units down.
6 - 8 ==> -2
Then,
range is (-2, ∞)
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