Radical Functions: Practice

Square both sides and rearrange.

$2x + 7 = x^2 + 4x + 4 \;\Rightarrow\; x^2 + 2x - 3 = 0 \;\Rightarrow\; (x + 3)(x - 1) = 0$

Testing the candidates, $x = 1$ gives $\sqrt{9} = 3 = 1 + 2$, while $x = -3$ gives $\sqrt{1} = 1 \ne -1$. Only $x = 1$ works.

Square both sides and factor.

$5x - 1 = x^2 + 2x + 1 \;\Rightarrow\; x^2 - 3x + 2 = 0 \;\Rightarrow\; (x - 1)(x - 2) = 0$

Testing both, $x = 1$ gives $\sqrt{4} = 2 = 1 + 1$ and $x = 2$ gives $\sqrt{9} = 3 = 2 + 1$. Both candidates check, so there are two solutions.

Match to $y = a\sqrt{x - h} + k$, so $h = -4$, $k = -2$, and $a = 3 > 0$.

$(h, k) = (-4, -2), \qquad x + 4 \ge 0 \Rightarrow x \ge -4$

The corner is $(-4, -2)$, the domain is $x \ge -4$, and since $a > 0$ the range is $y \ge -2$.

Square both sides and factor.

$x^2 = x + 6 \;\Rightarrow\; x^2 - x - 6 = 0 \;\Rightarrow\; (x - 3)(x + 2) = 0$

Since $x$ equals a square root it cannot be negative, so $x = -2$ is extraneous. Testing $x = 3$ gives $\sqrt{9} = 3$, so $x = 3$ is the only solution.

An $n$th root is a power of $\tfrac{1}{n}$, and a fourth root has index $4$.

$\sqrt[4]{x} = x^{1/4}$

The square root undoes squaring, so its inverse squares. The inverse's domain is the range of $\sqrt{x}$, which is $x \ge 0$.

$g(x) = x^2, \quad x \ge 0$

Restricting to $x \ge 0$ keeps $g$ a true inverse; allowing all real $x$ would fail the one-to-one requirement.

Both sides are single square roots, so squaring removes both at once.

$2x + 3 = x + 8 \;\Rightarrow\; x = 5$

Checking, both sides equal $\sqrt{13}$, so $x = 5$ works.

Both radicands must be nonnegative at the same time.

$x - 1 \ge 0 \text{ and } 5 - x \ge 0 \;\Rightarrow\; x \ge 1 \text{ and } x \le 5$

The overlap is $1 \le x \le 5$.

The radicand $-x$ must be nonnegative.

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

This is the base curve reflected across the y-axis, so it lives to the left of the origin, including $x = 0$.

The two graphs are inverses, so each is the reflection of the other across the line $y = x$, which swaps the coordinates.

$(a, b) \longrightarrow (b, a)$

Square both sides and factor.

$x - 3 = x^2 - 10x + 25 \;\Rightarrow\; x^2 - 11x + 28 = 0 \;\Rightarrow\; (x - 4)(x - 7) = 0$

Testing, $x = 7$ gives $\sqrt{4} = 2 = 7 - 5$, but $x = 4$ gives $\sqrt{1} = 1 \ne -1$. Only $x = 7$ survives.

The domain of $\sqrt{x - h}$ comes from $x - h \ge 0$, that is $x \ge h$.

$x \ge h = 6 \;\Rightarrow\; h = 6$