The problem defines a function $f(x, y) = x^2 y + \sqrt{y}$. We are asked to evaluate this function for different values of $x$ and $y$.

1. Problem Description

The problem defines a function $f(x, y) = x^2 y + \sqrt{y}$. We are asked to evaluate this function for different values of $x$ and $y$.

2. Solution Steps

(a) $f(2, 1)$:

Substitute $x = 2$ and $y = 1$ into the function:

$f(2, 1) = (2)^2 (1) + \sqrt{1} = 4(1) + 1 = 4 + 1 = 5$.

(b) $f(3, 0)$:

Substitute $x = 3$ and $y = 0$ into the function:

$f(3, 0) = (3)^2 (0) + \sqrt{0} = 9(0) + 0 = 0 + 0 = 0$.

(c) $f(1, 4)$:

Substitute $x = 1$ and $y = 4$ into the function:

$f(1, 4) = (1)^2 (4) + \sqrt{4} = 1(4) + 2 = 4 + 2 = 6$.

(d) $f(a, a^4)$:

Substitute $x = a$ and $y = a^4$ into the function:

$f(a, a^4) = (a)^2 (a^4) + \sqrt{a^4} = a^6 + a^2$ (assuming $a$ is non-negative).

(e) $f(1/x, x^4)$:

Substitute $x = 1/x$ and $y = x^4$ into the function:

$f(1/x, x^4) = (\frac{1}{x})^2 (x^4) + \sqrt{x^4} = \frac{1}{x^2} (x^4) + |x^2| = x^2 + x^2 = 2x^2$ (assuming $x$ is real).

(f) $f(2, -4)$:

Substitute $x = 2$ and $y = -4$ into the function:

$f(2, -4) = (2)^2 (-4) + \sqrt{-4}$.

Since the square root of a negative number is not a real number, the function is undefined for these values.

3. Final Answer

(a) $f(2, 1) = 5$

(b) $f(3, 0) = 0$

(c) $f(1, 4) = 6$

(d) $f(a, a^4) = a^6 + a^2$

(e) $f(1/x, x^4) = 2x^2$

(f) $f(2, -4)$ is undefined.