Interesting Mathematical Problems

Determining the domain of a function

What the domain of a function is and how to determine it in four cases: a product of roots, a fraction with a root and quadratic denominators, a difference of roots, and an expression with an absolute value.

The domain answers the question of which numbers may be substituted at all. Two conditions come back in every problem: what is under an even-degree root cannot be negative, and a denominator cannot be zero. When there are several conditions, the answer is their intersection.

Determine the domain of the function

What is the domain of a function?

The set of allowable inputs for which the function has a defined numerical value.

The inputs are most commonly denoted by x.

\sqrt{a} \;\Rightarrow\; a \geq 0

note that

a)

f(x) = \sqrt{4 - x} \cdot \sqrt{x + 1}

4 - x \geq 0

x \leq 4

x + 1 \geq 0

x \geq -1

x \in \langle -1, 4 \rangle

the intersection

b)

f(x) = \frac{\sqrt{x - 1}}{x^{2} - 2} + \frac{1}{x^{2} - 9}

x - 1 \geq 0

x \geq 1

x^{2} - 2 \neq 0

\left(x - \sqrt{2}\right)\left(x + \sqrt{2}\right) \neq 0

x \neq \sqrt{2}, \quad x \neq -\sqrt{2}

x^{2} - 9 \neq 0

(x - 3)(x + 3) \neq 0

x \neq 3, \quad x \neq -3

a^{2} - b^{2} = (a - b)(a + b)

x \in \langle 1, \sqrt{2}) \cup (\sqrt{2}, 3) \cup (3, +\infty)

the intersection

c)

f(x) = \sqrt{-x} - \sqrt{2 - x}

-x \geq 0

x \leq 0

2 - x \geq 0

x \leq 2

x \in (-\infty, 0 \rangle

the intersection

d)

f(x) = \frac{\sqrt{3 - 0.5x}}{\sqrt{4 + x}} - \frac{3}{|x| - 3}

3 - 0.5x \geq 0

0.5x \leq 3 \;/\cdot 2

x \leq 6

4 + x > 0

x > -4

|x| - 3 \neq 0

|x| \neq 3

x \neq -3, \quad x \neq 3

|3| = 3

|-3| = 3

x \in (-4, -3) \cup (-3, 3) \cup (3, 6 \rangle

the intersection

Frequently asked questions

What is the domain of a function?

It is the set of allowable inputs, that is the numbers for which the function has a defined numerical value. The inputs are most commonly denoted by x.

Why can a number under a root not be negative?

For an even-degree root no such value exists among the real numbers, because no square is negative. Hence the condition: the expression under the root must be greater than or equal to zero.

What do you do when there are several conditions?

Write each one out separately and at the end take their intersection - a number has to satisfy all of them at once, not just one.

How do you handle an absolute value in a denominator?

The denominator cannot be zero, so the absolute value of x cannot equal three. Since the modulus of three and of minus three give the same number, both are excluded: x not equal to 3 and x not equal to minus 3.

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