Radical Functions: Practice

Set the radicand nonnegative, and remember that dividing by a positive keeps the direction.

$9 - 3x \ge 0 \;\Rightarrow\; 9 \ge 3x \;\Rightarrow\; x \le 3$

Square both sides to clear the root, then solve.

$x - 2 = 16 \;\Rightarrow\; x = 18$

Checking, $\sqrt{18 - 2} = \sqrt{16} = 4$.

A power of $\tfrac{1}{3}$ is a cube root.

$27^{1/3} = \sqrt[3]{27} = 3$

The base range $y \ge 0$ is shifted down by $4$.

$y \ge 0 - 4 = -4$

So the outputs start at $-4$ and grow.

Square both sides, then solve the linear equation.

$3x + 4 = 25 \;\Rightarrow\; 3x = 21 \;\Rightarrow\; x = 7$

Checking, $\sqrt{3(7) + 4} = \sqrt{25} = 5$.

The radicand must be nonnegative, so the domain is $x - 5 \ge 0$, that is $x \ge 5$.

$4 < 5 \;\Rightarrow\; \sqrt{4 - 5} = \sqrt{-1} \text{ is not real}$

Every other choice is at least $5$, and $5$ itself gives $\sqrt{0} = 0$.

Test each candidate in the original equation $\sqrt{x} = x - 6$.

$\sqrt{9} = 3 = 9 - 6 \checkmark, \qquad \sqrt{4} = 2 \ne 4 - 6 = -2$

Only $x = 4$ fails, so it is the extraneous solution.

Take the root, then multiply by $2$.

$f(9) = 2\sqrt{9} = 2 \cdot 3 = 6$

An odd root of a negative number is real, since a negative cubed is negative.

$(-4)^3 = -64 \;\Rightarrow\; \sqrt[3]{-64} = -4$

Match to $y = a\sqrt{x - h} + k$. The inside $x + 3$ gives $h = -3$ and the constant is $k = -1$.

$(h, k) = (-3, -1)$

Square both sides, then solve for $x$.

$2x + 1 = 9 \;\Rightarrow\; 2x = 8 \;\Rightarrow\; x = 4$

Checking, $\sqrt{2(4) + 1} = \sqrt{9} = 3$.

The radicand must be nonnegative, giving a quadratic inequality.

$4 - x^2 \ge 0 \;\Rightarrow\; x^2 \le 4 \;\Rightarrow\; -2 \le x \le 2$