# 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
$(-\infty, -2) \cup (-2, 0) \cup (0, 1) \cup (1, +\infty)$