# Solutions to Domain of a Function Questions

Detailed solutions to the problems in Find the Domain of a Function - Problems are presented here.

## Solution to Problems

### Solution to Problem 1:

The given function is as follows
\[ f(x) = x + 1 \]
This a linear (polynomial) function and its domain is.

\( (-\infty, +\infty) \)

### Solution to Problem 2:

The given function is as follows
\[ f(x) = \sqrt{2x} \]
This a composed square root function. The domain is found by solving the inequality

\( 2x \geq 0 \)
The solution set for the above inequality is the domain and is given by the interval

\[ [0 , +\infty) \]

### Solution to Problem 3:

The given function is a rational function.
\[ f(x) = \frac{x - 1}{x - 3} \]
Its domain is the set of all real numbers except those values of x that make the denominator zero. Hence the domain is given by the interval

\[ (-\infty, 3) \cup (3, +\infty) \]

### Solution to Problem 4:

Find the domain of function f given by
\[ f(x) = \frac{\sqrt{-x + 1}}{x + 3} \]
To find the domain of the above function we need two conditions.

condition (1): \( -x + 1 \) is under the square root and must be positive or zero. Hence

\( -x + 1 \geq 0 \) leads to \( x \leq 1 \)

condition (2): \( x + 3 \) is in the denominator and must be non-zero. Hence x must not take the value -3. The two conditions must be satisfied simultaneously; hence the domain of the given function is given by

\[ (-\infty, -3) \cup (-3, 1] \]

### Solution to Problem 5:

Find the domain of function f given by.

\( f(x) = \sqrt[3]{2x + 1} \)
The expression \( 2x + 1 \) can take any real value. Hence the domain of the function is defined by

\[ (-\infty, +\infty) \]

### Solution to Problem 6:

The given function is

\( f(x) = \ln (x^2 - 9) \)
The expression \( x^2 - 9 \) must be positive for the function to be real valued. Hence we need to solve

\( x^2 - 9 > 0 \)

The above inequality can be solved by first factoring the left side.

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

The solution set to the above polynomial inequality, which also the domain of function f, is defined by.

\[ (-\infty, -3) \cup (3, +\infty) \]

### Solution to Problem 7:

The given function is
\[ f(x) = 2 \sin(x - 1) \]
\( x - 1 \) can be any real number. Hence the domain of the above function is given by
\[ (-\infty, +\infty) \]

### Solution to Problem 8:

Find the domain of function f defined by.
\[ f(x) = e^{(x - 4)} \]
\( x - 4 \) can take any real value and therefore the domain of \( f \) is the set of all real numbers or in interval form
\[ (-\infty, +\infty) \]

### Solution to Problem 9:

The given function is.
\[ f(x) = \arcsin(x^2 - 1) \]
For f to be real valued, the value of the expression \( x^2 - 1 \) must be restricted as follows:

\( -1 \leq x^2 - 1 \leq 1 \), (domain of arcsin function)
Solve the above inequality to obtain the solution set which also the domain of f.
\[ [- \sqrt{2}, \sqrt{2}]\]

### Solution to Problem 10:

Find the domain of.
\[ f(x) = \frac{1}{x^3 + x^2 -2x} \]
The domain of f is restricted to those values that do not make the denominator equal to zero. Let us find the values of x that make the denominator equal to zero.

\( x^3 + x^2 -2x = 0 \)

Factor the left side.

\( x ( x^2 + x -2) = 0 \)

The solutions to the above equations are:

\( x = 0 \), \( x = 1 \), and \( x = -2 \).

The domain of f is given by

\[ (-\infty, -2) \cup (-2, 0) \cup (0, 1) \cup (1, +\infty) \]

## Links and References

Math Problems, Questions and Online Self Tests.