# FIND THE DOMAIN AND RANGE OF AN ABSOLUTE VALUE FUNCTION ALGEBRAICALLY

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 upVertex is (9, 0)Range (0, ∞) y = -|9 - t|Domain  (-∞, ∞)y = -|t - 9|a = -1 opens downVertex 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) = | x2 + 3 |

Solution:

Domain:

A set of all defined value of  x is known as domain.

D(x) = x ∈ R

Domain: (-∞, ∞)

Range:

f(x) = | x2 + 3 |

Vertex is at (0, 3). The curve will open up. So, the range

Range: [3, ∞)

Problem 8 :

g(x) = | x2 + 3 | - 4

Solution:

Domain:

A set of all defined value of  x is known as domain.

D(x) = x ∈ R

Domain: (-∞, ∞)

Range:

g(x) = | x2 + 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| x2 + 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) = | t2 + 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) = | t2 + 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 |t2 + 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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