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