We are given a function of two variables, $f(x, y) = x^2y + \sqrt{y}$. We need to find the values of the function for the following input pairs: (a) $f(2, 1)$ (b) $f(3, 0)$ (c) $f(1, 4)$ (d) $f(a, a^4)$ (e) $f(1/x, x^4)$ (f) $f(2, -4)$

1. Problem Description

We are given a function of two variables, $f(x, y) = x^2y + \sqrt{y}$. We need to find the values of the function for the following input pairs:

(a) $f(2, 1)$

(b) $f(3, 0)$

(c) $f(1, 4)$

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

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

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

2. Solution Steps

The function is defined as:

$f(x, y) = x^2y + \sqrt{y}$

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

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

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

(d) $f(a, a^4) = (a)^2(a^4) + \sqrt{a^4} = a^6 + a^2$

Note: we assume $a^2 \geq 0$, so $\sqrt{a^4} = a^2$

(e) $f(1/x, x^4) = (1/x)^2(x^4) + \sqrt{x^4} = (1/x^2)(x^4) + x^2 = x^2 + x^2 = 2x^2$

Note: we assume $x^2 \geq 0$, so $\sqrt{x^4} = x^2$

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

Since we can only take the square root of a non-negative real number, $f(2, -4)$ is undefined in the real number domain.

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.