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

Problem: Which of these functions is even?

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

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Solution:

A function \( f \) is an even function if \( f(-x) = f(x) \) for all \( x \) in its domain.

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

Problem: 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. \( |x| - |y| = 0 \)

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Solution:

1. Solve for \( y \): \[ y = 5 - x^2 \] For each value of \( x \), there is exactly one corresponding value of \( y \). Therefore, \( y \) is a function of \( x \).
2. Solve for \( y \): \[ y^2 = 7x - x^2 \implies y = \pm\sqrt{7x - x^2} \] There are values of \( x \) that yield two distinct values of \( y \). Therefore, \( y \) is not a function of \( x \).
3. Solve for \( y \): \[ y^3 = x^2 - 9 \implies y = \sqrt[3]{x^2 - 9} \] For each value of \( x \), there is one unique real value of \( y \). Therefore, \( y \) is a function of \( x \).
4. Solve for \( y \): \[ |y| = |x| \implies y = \pm x \] There are values of \( x \) that yield two values of \( y \). Therefore, \( y \) is not a function of \( x \).

Problem: 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. \( D = \{ (a, 5), (a, -9), (a, 0), (a, 12) \} \)

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Solution:

1. For each input \( x \), there corresponds exactly one output \( y \). Hence, Relation A is a function.
2. The pairs \((b, 6)\) and \((b, 9)\) mean that for \( x = b \), there are two different outputs (6 and 9). Hence, Relation B is not a function.
3. Each input maps to a single output value. Hence, Relation C is a function.
4. For input \( x = a \), there are multiple outputs (5, -9, 0, 12). Hence, Relation D is not a function.

Problem: 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. \( j(x) = \dfrac{1}{(x + 1)(x - 7)} \)

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Solution:

1. \( f(x) = \dfrac{1}{x - 2} \): Division by zero is not allowed, so \( x - 2 \neq 0 \implies x \neq 2 \).
   Domain: \( (-\infty, 2) \cup (2, +\infty) \)
2. \( g(x) = \sqrt{x + 5} \): The expression under the square root must be non-negative: \( x + 5 \geq 0 \implies x \geq -5 \).
   Domain: \( [-5, +\infty) \)
3. \( h(x) = \dfrac{3}{\sqrt{x - 4}} \): The expression under the square root must be strictly positive (denominator cannot be zero): \( x - 4 > 0 \implies x > 4 \).
   Domain: \( (4, +\infty) \)
4. \( j(x) = \dfrac{1}{(x + 1)(x - 7)} \): Denominator cannot be zero, so \( x \neq -1 \) and \( x \neq 7 \).
   Domain: \( (-\infty, -1) \cup (-1, 7) \cup (7, +\infty) \)

Problem: 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. \( k(x) = x^2 + 3 \)

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Solution: The zeros of a function are the solutions to \( f(x) = 0 \).

1. \( 3 - \dfrac{5}{x} = 0 \implies \dfrac{5}{x} = 3 \implies x = \dfrac{5}{3} \)
   Zero at \( x = \dfrac{5}{3} \)
2. \( \sqrt{x - 1} = 0 \implies x - 1 = 0 \implies x = 1 \)
   Zero at \( x = 1 \)
3. \( |x - 9| - 4 = 0 \implies |x - 9| = 4 \implies x - 9 = \pm 4 \)
   \( x = 13 \) or \( x = 5 \)
   Zeros at \( x = 5 \) and \( x = 13 \)
4. \( |x - 9| + 7 = 0 \implies |x - 9| = -7 \)
   Since an absolute value cannot be negative, there are no zeros.
5. \( x^2 - 16 = 0 \implies x^2 = 16 \implies x = \pm 4 \)
   Zeros at \( x = -4 \) and \( x = 4 \)
6. \( x^2 + 3 = 0 \implies x^2 = -3 \)
   Since \( x^2 \geq 0 \) for all real numbers, there are no real zeros.

Problem: Find the range of each function.

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

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Solution:

1. \( f(x) = x^2 \): The square of any real number is non-negative: \( x^2 \geq 0 \).
   Range: \( [0, +\infty) \)
2. \( g(x) = |x| \): Absolute values are always non-negative: \( |x| \geq 0 \).
   Range: \( [0, +\infty) \)
3. \( h(x) = x^2 + 6 \): Since \( x^2 \geq 0 \), adding 6 gives \( x^2 + 6 \geq 6 \).
   Range: \( [6, +\infty) \)
4. \( j(x) = |x| + 2 \): Since \( |x| \geq 0 \), adding 2 gives \( |x| + 2 \geq 2 \).
   Range: \( [2, +\infty) \)

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

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

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Solution:

1. \( f(2) = 2(2)^2 - 4(2) + 4 = 2(4) - 8 + 4 = 4 \)
2. \( h(5) = 2(5) - 4 = 10 - 4 = 6 \)
3. \( f(3) + h(3) = [2(3)^2 - 4(3) + 4] + [2(3) - 4] = [18 - 12 + 4] + [6 - 4] = 10 + 2 = 12 \)

Problem: 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. \( |x - 3| + |5 - y| = 6 \)

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Solution:

1. \( 2x + 4y = 5 \)
   Set \( y = 0 \): \( 2x = 5 \implies x = \dfrac{5}{2} \) → x-intercept: \( \left(\dfrac{5}{2}, 0\right) \)
   Set \( x = 0 \): \( 4y = 5 \implies y = \dfrac{5}{4} \) → y-intercept: \( \left(0, \dfrac{5}{4}\right) \)
2. \( x^2 + (y - 3)^2 = 9 \)
   Set \( y = 0 \): \( x^2 + 9 = 9 \implies x = 0 \) → x-intercept: \( (0, 0) \)
   Set \( x = 0 \): \( (y - 3)^2 = 9 \implies y - 3 = \pm 3 \implies y = 0 \text{ or } y = 6 \) → y-intercepts: \( (0, 0) \) and \( (0, 6) \)
3. \( |x - 3| + |5 - y| = 6 \)
   Set \( y = 0 \): \( |x - 3| + 5 = 6 \implies |x - 3| = 1 \implies x - 3 = \pm 1 \implies x = 4 \text{ or } x = 2 \) → x-intercepts: \( (4, 0) \) and \( (2, 0) \)
   Set \( x = 0 \): \( 3 + |5 - y| = 6 \implies |5 - y| = 3 \implies 5 - y = \pm 3 \implies y = 2 \text{ or } y = 8 \) → y-intercepts: \( (0, 2) \) and \( (0, 8) \)

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

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Solution:

Substitute the given conditions into \( f(x) = Ax + B \):

\[ \begin{cases} 2A + B = 1 \\ 4A + B = -3 \end{cases} \]

Subtract the first equation from the second:

\[ (4A + B) - (2A + B) = -3 - 1 \implies 2A = -4 \implies A = -2 \]

Substitute \( A = -2 \) into the first equation:

\[ 2(-2) + B = 1 \implies -4 + B = 1 \implies B = 5 \]

Final Answer: \[ f(x) = -2x + 5 \]

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

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Solution:

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

\[ \dfrac{7 - 3}{4 - 2} = \dfrac{b - 7}{8 - 4} \] \[ \dfrac{4}{2} = \dfrac{b - 7}{4} \implies 2 = \dfrac{b - 7}{4} \]

Solve for \( b \):

\[ 8 = b - 7 \implies b = 15 \]