College Algebra Questions With Answers
Domain and Range of Functions

The domain is found by setting the denominator not equal to zero because division by 0 is not allowed:

\[ |x^2 - 4| \neq 0 \implies x^2 - 4 \neq 0 \implies x^2 \neq 4 \implies x \neq \pm 2 \]

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

The denominator cannot be zero:

\[ x^2 + 4x + 3 \neq 0 \]

Solve the equation \( x^2 + 4x + 3 = 0 \) by factoring:

\[ (x + 3)(x + 1) = 0 \implies x = -3 \text{ and } x = -1 \]

Thus, \( x \) must not equal \(-3\) or \(-1\).

Domain: \((- \infty, -3) \cup (-3, -1) \cup (-1, +\infty)\)

The expression under the square root must be greater than or equal to zero for the function to be real:

\[ x^2 + 5x - 6 \geq 0 \]

Factor the quadratic expression:

\[ (x + 6)(x - 1) \geq 0 \]

Solving this inequality gives the solution set:

Domain: \((- \infty, -6] \cup [1, +\infty)\)

The quantity under the square root must be positive because it is in the denominator and division by zero is not allowed:

\[ (x - 2)^2 > 0 \]

Since it is a squared term, it is always non-negative, so it cannot be zero when \( x \neq 2 \).

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

We must satisfy two conditions:

• The expression under the square root must be non-negative: \( x + 2 \geq 0 \implies x \geq -2 \).
• The denominator cannot be zero: \( x - \sqrt{x + 2} \neq 0 \).

Solve \( x = \sqrt{x + 2} \):

\[ x^2 = x + 2 \implies x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0 \]

This gives potential solutions \( x = 2 \) and \( x = -1 \). Checking \( x = -1 \): \(-1 - \sqrt{1} = -2 \neq 0\) (valid). Checking \( x = 2 \): \(2 - \sqrt{4} = 0\) (division by zero, so \( x = 2 \) must be excluded).

Domain: \([-2, 2) \cup (2, +\infty)\)

The argument of the logarithm must be strictly positive for the function to take real values:

\[ |x + 3| > 0 \]

This is true for all real numbers except when \( x = -3 \).

Domain: \((- \infty, -3) \cup (-3, +\infty)\)

Rewrite the quadratic function in vertex form by completing the square:

\[ f(x) = -(x^2 - 6x) + 5 = -(x - 3)^2 + 9 + 5 = -(x - 3)^2 + 14 \]

Since \( -(x - 3)^2 \leq 0 \) for all real numbers \( x \), adding 14 gives:

\[ -(x - 3)^2 + 14 \leq 14 \]

Range: \((- \infty, 14]\)

The absolute value function is always greater than or equal to zero:

\[ |x + 3| \geq 0 \]

Subtracting 2 from both sides gives:

\[ |x + 3| - 2 \geq -2 \]

Range: \([-2, +\infty)\)

To find the range of \( h(x) \), find the domain of its inverse function \( h^{-1}(x) \):

\[ y = \frac{x - 2}{x + 3} \implies y(x + 3) = x - 2 \implies xy + 3y = x - 2 \] \[ xy - x = -3y - 2 \implies x(y - 1) = -3y - 2 \implies x = \frac{-3y - 2}{y - 1} \]

Thus, the inverse function is:

\[ h^{-1}(x) = \frac{-3x - 2}{x - 1} \]

The domain of \( h^{-1}(x) \) requires \( x \neq 1 \).

Range: \((- \infty, 1) \cup (1, +\infty)\)

The inner expression \( x^3 + 4 \) has a range of \((- \infty, +\infty)\). Because of the absolute value, all negative outputs are reflected to positive values, and zero is included.

Range: \([0, +\infty)\)

The graph of \( y = (x + 4)(x - 2) \) is a parabola opening upward with \( x \)-intercepts at \( x = -4 \) and \( x = 2 \), and a vertex at \((-1, -9)\). The portion of the graph between the \( x \)-intercepts lies below the \( x \)-axis (negative values). Applying the absolute value reflects this negative portion above the \( x \)-axis.

Range: \([0, +\infty)\)

The standard function \( \dfrac{1}{x} \) has a range of \((- \infty, 0) \cup (0, +\infty)\). Applying absolute value changes negative outputs to positive, resulting in \((0, +\infty)\). Shifting the graph horizontally by 3 units does not alter its vertical range.

Range: \((0, +\infty)\)