Intermediate Algebra Practice: Functions, Domain,
Range & Even Functions with Solutions - Sample 4

This page offers a collection of intermediate algebra problems covering key topics such as functions, domain, range, collinear points, zeros of functions, and x- and y-intercepts, complete with detailed answers. Solutions and detailed explanatiosn are also provided.

Questions 1

Which of these functions is even?

  1. \( f(x) = -x^2 + 7 \)
  2. \( g(x) = |x - 6| \)
  3. \( h(x) = x^3 + 9 \)
  4. \( j(x) = |x| + 1 \)

Solution:

If a given function \( f \) is such that \( f(-x) = f(x) \), then \( f \) is an even function.

  1. \( f(x) = -x^2 + 7 \),
    \( f(-x) = -(-x)^2 + 7 = -x^2 + 7 \).
    Function \( f \) is an even function.
  2. \( g(x) = |x - 6| \),
    \( g(-x) = |-x - 6| = |-(x + 6)| = |x + 6| \).
    Function \( g \) is not even.
  3. \( h(x) = x^3 + 9 \),
    \( h(-x) = (-x)^3 + 9 = -x^3 + 9 \).
    Function \( h \) is not even.
  4. \( j(x) = |x| + 1 \),
    \( j(-x) = |-x| + 1 = |x| + 1 \).
    Function \( j \) is even.

Questions 2

Which of the following equations represent \( y \) as a function of \( x \)?

  1. \( x^2 + y = 5 \)
  2. \( x^2 - 7x + y^2 = 0 \)
  3. \( x^2 - y^3 - 9 = 0 \)
  4. \( |x| - |y| = 0 \)

Solution:

  1. Solve for \( y \) to find: \[ y = 5 - x^2 \] For each value of \( x \), there is only one corresponding value of \( y \). Therefore, \( y \) is a function of \( x \).
  2. Solve for \( y \) to find: \[ y = \pm \sqrt{7x - x^2} \] There are values of \( x \) that give two values of \( y \) and therefore \( y \) is not a function of \( x \).
  3. Solve for \( y \) to find: \[ y^3 = (x^2 - 9)^{\dfrac{1}{3}} \] For each value of \( x \), there is one value of \( y \) and therefore \( y \) is a function of \( x \).
  4. Solve for \( y \) to find: \[ y = \pm x \] There are values of \( x \) that give two values of \( y \) and therefore \( y \) is not a function of \( x \).

Questions 3

Which set of ordered pairs represents a function?

  1. \( A = \{ (a , 3), (b , 5), (c , 9), (d , 9) \} \)
  2. \( B = \{ (a , -3), (b , 6), (c , 1), (b , 9) \} \)
  3. \( C = \{ (a , 3), (b , 3), (c , 3), (d , 3) \} \)
  4. \( D = \{ (a , 5), (a , -9), (a , 0), (a , 12) \} \)

Solution:

  1. For each \( x \) (first value) there corresponds only one value of \( y \) (second value), hence Relation A is a function.

  2. The two pairs \((b, 6)\) and \((b, 9)\) mean that for \( x = b \), there are two different values of \( y \): 6 and 9, hence Relation B is not a function.

  3. For each \( x \) value there corresponds only one value of \( y \), hence Relation C is a function.

  4. For \( x = a \), \( y \) has different values, hence Relation D is not a function.

Questions 4

Find the domain of each function.

  1. \( f(x) = \dfrac{1}{x - 2} \)
  2. \( g(x) = \sqrt{x + 5} \)
  3. \( h(x) = \dfrac{3}{\sqrt{x - 4}} \)
  4. \( j(x) = \dfrac{1}{(x + 1)(x - 7)} \)

Solution:

  1. \( f(x) = \dfrac{1}{x - 2} \), division by zero is not allowed, hence \( x - 2 \neq 0 \).

    Hence the Domain: \((-\infty, 2) \cup (2, +\infty)\)

  2. \( g(x) = \sqrt{x + 5} \), for \( g(x) \) to be real, \( x \) must satisfy the inequality: \[ x + 5 \geq 0 \]. which may be written ad: \[ x \geq -5 \] Hence the Domain: \([ -5, +\infty )\)

  3. \( h(x) = \dfrac{3}{\sqrt{x - 4}} \), for \( h(x) \) to be real, \( x \) must satisfy the inequality: \( x - 4 > 0 \). which may be written ad: \[ x \ge 4 \] Hence the Domain: \((4, +\infty)\)

  4. \( j(x) = \dfrac{1}{(x + 1)(x - 7)} \), division by zero is not allowed, hence \( x \neq -1 \) and \( x \neq 7 \).

    Domain: \((-\infty, -1) \cup (-1, 7) \cup (7, +\infty)\)

Questions 5

Find the zeros of each function.

  1. \( f(x) = 3 - \dfrac{5}{x} \)
  2. \( g(x) = \sqrt{x - 1} \)
  3. \( h(x) = |x - 9| - 4 \)
  4. \( i(x) = |x - 9| + 7 \)
  5. \( j(x) = x^2 - 16 \)
  6. \( k(x) = x^2 + 3 \)

Solution:

The zeros of a given function f are the solution(s) to the equation \( f(x) = 0 \).

  1. \( f(x) = 3 - \dfrac{5}{x} \)
    Set \( f(x) = 0 \): \[ 3 - \dfrac{5}{x} = 0 \quad \Rightarrow \quad \dfrac{5}{x} = 3 \quad \Rightarrow \quad x = \dfrac{5}{3} \] Zero at \( x = \dfrac{5}{3} \)
  2. \( g(x) = \sqrt{x - 1} \)
    Set \( g(x) = 0 \): \[ \sqrt{x - 1} = 0 \quad \Rightarrow \quad x - 1 = 0 \quad \Rightarrow \quad x = 1 \] Zero at \( x = 1 \)
  3. \( h(x) = |x - 9| - 4 \)
    Set \( h(x) = 0 \): \[ |x - 9| - 4 = 0 \quad \Rightarrow \quad |x - 9| = 4 \] This gives: \[ x - 9 = 4 \Rightarrow x = 13, \quad x - 9 = -4 \Rightarrow x = 5 \] Zeros at \( x = 5 \) and \( x = 13 \)
  4. \( i(x) = |x - 9| + 7 \)
    Set \( i(x) = 0 \): \[ |x - 9| + 7 = 0 \quad \Rightarrow \quad |x - 9| = -7 \] No solution, since absolute value cannot be negative. No zeros
  5. \( j(x) = x^2 - 16 \)
    Set \( j(x) = 0 \): \[ x^2 - 16 = 0 \quad \Rightarrow \quad x^2 = 16 \quad \Rightarrow \quad x = \pm 4 \] Zeros at \( x = -4 \) and \( x = 4 \)
  6. \( k(x) = x^2 + 3 \)
    Set \( k(x) = 0 \): \[ x^2 + 3 = 0 \quad \Rightarrow \quad x^2 = -3 \] No real solution since \( x^2 \geq 0 \) for all real numbers. No real zeros

Questions 6

Find the range of each function.

  1. \( f(x) = x^2 \)
  2. \( g(x) = |x| \)
  3. \( h(x) = x^2 + 6 \)
  4. \( j(x) = |x| + 2 \)

Solution:

The range of a given function \( f \) is the set of values of \( f(x) \) for all \( x \) in the domain of \( f \).

  1. \( f(x) = x^2 \).
    The square of any real number \( x \) is either positive or zero, with equality when \( x = 0 \). Therefore, \[ x^2 \geq 0 \Rightarrow f(x) \geq 0. \] Range: \( [0, +\infty) \)
  2. \( g(x) = |x| \).
    The absolute value of a real number is always non-negative, with equality when \( x = 0 \). Therefore, \[ |x| \geq 0 \Rightarrow g(x) \geq 0. \] Range: \( [0, +\infty) \)
  3. \( h(x) = x^2 + 6 \).
    Since \( x^2 \geq 0 \), adding 6 to both sides gives: \[ x^2 + 6 \geq 6 \Rightarrow h(x) \geq 6. \] Range: \( [6, +\infty) \)
  4. \( j(x) = |x| + 2 \).
    Since \( |x| \geq 0 \), adding 2 to both sides gives: \[ |x| + 2 \geq 2 \Rightarrow j(x) \geq 2. \] Range: \( [2, +\infty) \)

Questions 7

Let \( f(x) = 2x^2 - 4x + 4 \) and \( h(x) = 2x - 4 \). Evaluate:

  1. \( f(2) \)
  2. \( h(5) \)
  3. \( f(3) + h(3) \)

Solution:

Let \( f(x) = 2x^2 - 4x + 4 \) and \( h(x) = 2x - 4 \).

  1. \( f(2) = 2(2)^2 - 4(2) + 4 = 4 \)
  2. \( h(5) = 2(5) - 4 = 6 \)
  3. \( f(3) + h(3) = 2(3)^2 - 4(3) + 4 + 2(3) - 4 = 12 \)

Questions 8

Find the x and y intercepts of the graphs of the following equations.

  1. \( 2x + 4y = 5 \)
  2. \( x^2 + (y - 3)^2 = 9 \)
  3. \( |x - 3| + |5 - y| = 6 \)

Solution:

  1. \( 2x + 4y = 5 \). To find the x-intercept, set \( y = 0 \) in the given equation and solve for \( x \).

    \( 2x + 4(0) = 5 \Rightarrow x = \dfrac{5}{2} \), so the x-intercept is \( \left( \dfrac{5}{2}, 0 \right) \).

    To find the y-intercept, set \( x = 0 \) in the equation and solve for \( y \).

    \( 2(0) + 4y = 5 \Rightarrow y = \dfrac{5}{4} \), so the y-intercept is \( \left( 0, \dfrac{5}{4} \right) \).
  2. \( x^2 + (y - 3)^2 = 9 \)

    Set \( y = 0 \) and solve for \( x \): \( x^2 + (0 - 3)^2 = 9 \Rightarrow x^2 + 9 = 9 \Rightarrow x = 0 \)
    x-intercept: \( (0, 0) \)

    Set \( x = 0 \) and solve for \( y \): \( 0^2 + (y - 3)^2 = 9 \Rightarrow (y - 3)^2 = 9 \Rightarrow y - 3 = \pm3 \Rightarrow y = 0 \) or \( y = 6 \)
    y-intercepts: \( (0, 0) \) and \( (0, 6) \)
  3. \( |x - 3| + |5 - y| = 6 \)

    Set \( y = 0 \): \( |x - 3| + |5 - 0| = 6 \Rightarrow |x - 3| + 5 = 6 \Rightarrow |x - 3| = 1 \Rightarrow x - 3 = \pm1 \Rightarrow x = 4 \) or \( x = 2 \)
    x-intercepts: \( (4, 0) \) and \( (2, 0) \)

    Set \( x = 0 \): \( |0 - 3| + |5 - y| = 6 \Rightarrow 3 + |5 - y| = 6 \Rightarrow |5 - y| = 3 \Rightarrow 5 - y = \pm3 \Rightarrow y = 2 \) or \( y = 8 \)
    y-intercepts: \( (0, 2) \) and \( (0, 8) \)

Questions 9

Find the linear function \( f(x) = A x + B \) such that \( f(2) = 1 \) and \( f(4) = -3 \).

Solution:

We are given the function: \( f(x) = Ax + B \), and the values \( f(2) = 1 \) and \( f(4) = -3 \).

Using \( f(2) = 1 \), we substitute into the function: \[ 1 = 2A + B \] Using \( f(4) = -3 \), we substitute into the function: \[ -3 = 4A + B \] We now solve the system of equations: \[ \begin{cases} 1 = 2A + B \\ -3 = 4A + B \end{cases} \] Subtract the first equation from the second: \[ (-3) - 1 = (4A + B) - (2A + B) \Rightarrow -4 = 2A \Rightarrow A = -2 \] Substitute \( A = -2 \) into the first equation: \[ 1 = 2(-2) + B \Rightarrow 1 = -4 + B \Rightarrow B = 5 \] Therefore, the function is: \[ f(x) = -2x + 5 \]

Questions 10

Find the constant \( b \) so that the three points \( A(2, 3),\ B(4, 7), \text{ and } C(8, b) \) are collinear (i.e., lie on the same line).

Solution:

The slope of the line through points \( A \) and \( B \) must be equal to the slope of the line through points \( B \) and \( C \).

Use the slope equality condition:
\[ \dfrac{7 - 3}{4 - 2} = \dfrac{b - 7}{8 - 4} \] Simplify :
\[ 2 = \dfrac{b - 7}{4} \] Solve for \( b \): \[ 2 \times 4 = b - 7 \implies 8 = b - 7 \implies b = 15 \]

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