Radical Functions: Practice

The square root asks which nonnegative number, squared, gives $49$.

$7^2 = 49 \;\Rightarrow\; \sqrt{49} = 7$

Substitute $x = 36$ into the rule and take the principal root.

$f(36) = \sqrt{36} = 6$

The index is even, so the radicand cannot be negative. Set it to be nonnegative and solve.

$x - 4 \ge 0 \;\Rightarrow\; x \ge 4$

A square root needs a radicand that is zero or greater, and $\sqrt{0} = 0$ is allowed.

$x \ge 0$

The input $0$ is included, so the domain uses $\ge$, not a strict $>$.

The smallest allowed input is $x = 0$, and $\sqrt{0} = 0$.

$(0, \sqrt{0}) = (0, 0)$

The curve starts at the corner $(0, 0)$ and climbs to the right.

Take the root of $0$ first, then add $3$.

$f(0) = \sqrt{0} + 3 = 0 + 3 = 3$

Square both sides to undo the root.

$x = 5^2 = 25$

Checking, $\sqrt{25} = 5$, so $x = 25$ works.

The cube root asks which number, cubed, gives $8$.

$2^3 = 8 \;\Rightarrow\; \sqrt[3]{8} = 2$

Set the radicand to be nonnegative and solve.

$x + 7 \ge 0 \;\Rightarrow\; x \ge -7$

A real square is never negative, so no real number squares to give $-4$.

$\sqrt{-4} \text{ has no real value}$

The others are fine: $\sqrt{4} = 2$, the odd cube root $\sqrt[3]{-4}$ is real, and $\sqrt{0} = 0$.

Compute the radicand, then take the root.

$f(10) = \sqrt{10 - 1} = \sqrt{9} = 3$

Find the nonnegative number whose square is $0$.

$0^2 = 0 \;\Rightarrow\; \sqrt{0} = 0$

Zero is a valid input, so the root is defined and equals $0$.