Questions on Functions with Solutions

A set of questions on functions and their properties are presented. Detailed solutions to the questions are also provided.

Question 1:

\( f \) is a function defined by \[ f(x) = \begin{cases} 2x + 4 & x \leq 2 \\ 2x - 1 & x > 2 \end{cases} \] Find \( f(0), f(2) \) and \(f(4)\).

Question 2:

Is \( y \) a function of \( x \) in the following equation \[ -x^2 + y^2 - 2 = 0 \]

Question 3:

Find the domain of function f defined by \[ f(x) = \sqrt{x^2 + 1} + 2 \]

Question 4:

Let \( f \) be a function defined by \[ f(x) = |x - 1| + 3. \] Find \( f(-9) \).

Question 5:

Let \( f \) and \( g \) be functions defined by \[ f(x) = 2x + 2 \quad \text{and} \quad g(x) = \sqrt{x + 3}. \] Find \((f \circ g)(6)\).

Question 6:

Find all real values of \( x \) such that \( f(x) = 0 \), given that \( f \) is a function defined by \[ f(x) = \frac{x^2 + 2x - 3}{x - 1}. \]

Question 7:

Find all real values of \( x \) such that \( f(x) = g(x) \), where the functions \( f \) and \( g \) are given by: \[ f(x) = 3x + \sqrt{x}, \quad g(x) = 2x + 6 \]

Question 8:

Is \(y\) a function of \(x\) in the following equation? \[ |y + 1| = x^2 + 2 \]

Question 9:

Find the domain of the function \( f \) defined by \[ f(x) = \frac{1}{\sqrt{|x - 1|}} \]

Solutions to the Above Questions

Solution to Question 1:

Since \(0 \lt 2\), \(f(0)\) is found by substituting \(x = 0\) in the formula \( f(x) = 2x + 4 \) \[ f(0) = 2 \cdot 0 + 4 = 4 \] Since \(2 \le 2\), \(f(2)\) is found by substituting \(x = 2\) in the formula \[ f(x) = 2x + 4 \] \[ f(2) = 2 \cdot 2 + 4 = 8 \] Since \(4 > 2\), \(f(4)\) is found by substituting \(x = 4\) in the formula \[ f(x) = 2x - 1 \] \[ f(4) = 2 \cdot 4 - 1 = 7 \]

Solution to Question 2:

To answer the question, we need to solve the given equation for \(y\): \[ y = \pm \sqrt{x^2 + 2} \] \(y\) is not a function of \(x\) because for one value of the independent variable \(x\) (input) we obtain two values for the dependent variable \(y\) (output).

Solution to Question 3:

The expression inside the square root is positive for any real value of \( x \); therefore the domain of function \( f \) is the set of all real numbers.

Solution to Question 4:

Evaluate \( f(-9) \) by substituting \( x = -9 \) in the formula of the function: \[ f(-9) = |-9 - 1| + 3 = 10 + 3 = 13 \]

Solution to Question 5:

Calculate \( (f \circ g)(6) \) as follows: \[ (f \circ g)(6) = f(g(6)) \] Evaluate \( g(6)\): \[ g(6) = \sqrt{6 + 3} = 3 \] Substitute \( g(6) \) by \( 3 \) in \( f(g(6)) \): \[ f(g(6)) = f(3) \] Evaluate \( f(3) \): \[ f(3) = 2 \cdot 3 + 2 = 8 \] Thus, \[ (f \circ g)(6) = 8 \].

Solution to Question 6:

We have \( f(x) = 0 \), which gives the following equation: \[ \frac{x^2 + 2x - 3}{x - 1} = 0 \] For values of \( x \neq 1 \), the above equation leads to: \[ x^2 + 2x - 3 = 0 \] Solve the above equation by factoring: \[ (x - 1)(x + 3) = 0 \] This equation has two solutions: \[ x = 1 \quad \text{and} \quad x = -3 \] However, \( x = 1 \) is NOT a solution to \( f(x) = 0 \) because it makes the denominator of \( f(x) \) equal to zero. Therefore, the only value of \( x \) for which \( f(x) = 0 \) is: \[ x = -3 \]

Solution to Question 7:

We start with the equation \( g(x) = g(x) \), which leads to: \[ 3x + \sqrt{x} = 2x + 6 \] Rewriting the equation: \[ \sqrt{x} = -x + 6 \] Square both sides: \[ x = (-x + 6)^2 \] Simplify: \[ x = x^2 + 36 - 12x \] Rewrite in standard quadratic form: \[ x^2 - 13x + 36 = 0 \] Solve the quadratic equation: \[ x = 4 \quad \text{or} \quad x = 9 \] Since we squared both sides, extraneous solutions may have been introduced.
1) Check \( x = 4 \)
(left habd side of the original equation) LHS = \( 3(4) + \sqrt{4} = 14 \)
(Right habd side of the original equation) RHS = \( 2(4)+6 = 14 \)
\( x = 4 \) satisfies the original equation.
1) Check \( x = 9 \)
(left habd side of the original equation) LHS = \( 3(9) + \sqrt{9} = 30 \)
(Right habd side of the original equation) RHS = \( 2(9)+6 = 24 \)
\( x = 9 \) does NOT satisfies the original equation and therefore is not a solution to the original equation.
Conclusion: \( x = 4 \) is the only value for which \( g(x) = g(x) \).

Solution to Question 8:

Solve the given equation for \(y\): \[ y + 1 = \pm (x^2+2) \] which gives two solutions \[ y = -1 - (x^2 + 2) \quad \text{or} \quad y = -1 + (x^2 + 2) \] The two solutions mean that for one value of \(x\) we obtain two values of \(y\), and hence \(y\) is not a function of \(x\).

Solution to Question 9:

The domain of function \( f \) is the set of all real values of \( x \) for which \( f(x) \) is real.
\( |x - 1| \) is either positive or zero at \( x = 1 \). But \( |x - 1| \) is in the denominator. Therefore the domain of \( f \) is the set of all real numbers except \( 1 \).