Question: For the given function, find $f(22)$, $f(-13)$, $f(-130)$, and $f(3)$. $f(x) = \sqrt[3]{x + 5}$ Find $f(22)$. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. $f(22) = [ ]$ (Simplify your answer.) B. The root is not a real number.

1. **State the problem:** We need to find the values of the function $f(x) = \sqrt[3]{x + 5}$ at $x=22$, $x=-13$, $x=-130$, and $x=3$. 2. **Recall the formula:** The cube root function is defined as $f(x) = \sqrt[3]{x + 5} = (x+5)^{1/3}$. Cube roots are defined for all real numbers, so the root will always be a real number. 3. **Calculate $f(22)$:** $$f(22) = \sqrt[3]{22 + 5} = \sqrt[3]{27}$$ Since $27 = 3^3$, $$f(22) = 3$$ 4. **Calculate $f(-13)$:** $$f(-13) = \sqrt[3]{-13 + 5} = \sqrt[3]{-8}$$ Since $-8 = (-2)^3$, $$f(-13) = -2$$ 5. **Calculate $f(-130)$:** $$f(-130) = \sqrt[3]{-130 + 5} = \sqrt[3]{-125}$$ Since $-125 = (-5)^3$, $$f(-130) = -5$$ 6. **Calculate $f(3)$:** $$f(3) = \sqrt[3]{3 + 5} = \sqrt[3]{8}$$ Since $8 = 2^3$, $$f(3) = 2$$ **Final answers:** - $f(22) = 3$ - $f(-13) = -2$ - $f(-130) = -5$ - $f(3) = 2$