↔↕Domain and Range Practice Problems

A polynomial has nothing that can break, so the domain is all real numbers, \((-\infty,\ \infty)\).

The denominator is zero at \(x = 0\), so exclude it: \((-\infty,\ 0) \cup (0,\ \infty)\).

An even root needs a non-negative inside: \(x + 6 \ge 0\), so \(x \ge -6\), giving \([-6,\ \infty)\). Zero is allowed under a root, so the bracket is square.

Only the denominator matters. \(x - 7 = 0\) at \(x = 7\), so exclude it.

The smallest \(x^2\) can be is \(0\), so the smallest output is \(3\). Range \([3,\ \infty)\).

\(10 - x \ge 0\) gives \(-x \ge -10\), so \(x \le 10\). Dividing by \(-1\) reversed the inequality.

\(x^2 - 4 = 0\) at \(x = 2\) and \(x = -2\), so both are excluded, leaving three pieces.

A cube root is an odd root, and \(\sqrt[3]{-8} = -2\) is perfectly real. There is no restriction at all.

A logarithm needs a strictly positive argument: \(x - 5 > 0\), so \(x > 5\). The bracket is round because zero is not allowed.

Vertex form gives the vertex \((-2,\ -7)\), and the parabola opens upward, so \(-7\) is the minimum.

The negative in front means the parabola opens downward, so the vertex height \(9\) is a maximum.

Two conditions at once: the root needs \(x \ge 0\), and the denominator excludes \(x = 9\). Take the overlap.

The root needs \(9 - x \ge 0\), but the root is in a denominator so it cannot be zero either. That tightens it to \(9 - x > 0\), giving \(x < 9\).

\(\sqrt{x-2} \ge 0\), so the output is at least \(1\). Note the domain here is \([2,\ \infty)\) — domain and range are different intervals.

The vertex \(x = 0\) lies inside the interval, so the minimum is \(f(0) = 0\), not an endpoint value. The endpoints give \(f(-3) = 9\) and \(f(2) = 4\), so the maximum is \(9\). Range \([0,\ 9]\).