Question

match the functions with the graphs of their domains.

1. $f(x,y)=sqrt{x^{3}y^{5}}$
2. $f(x,y)=ln(x + 2y)$
3. $f(x,y)=x + 2y$
4. $f(x,y)=e^{\frac{1}{x+2y}}$

a.
b.
c.

Explanation:

Step1: Find domain of $f(x,y)=\sqrt{x^{3}y^{5}}$

For the square - root function, the expression inside the square - root must be non - negative, i.e., $x^{3}y^{5}\geq0$. This implies that either $x\geq0,y\geq0$ or $x\leq0,y\leq0$. The domain is the union of the first and third quadrants.

Step2: Find domain of $f(x,y)=\ln(x + 2y)$

The argument of the natural logarithm function must be positive. So $x+2y>0$, which represents the region above the line $y=-\frac{1}{2}x$.

Step3: Find domain of $f(x,y)=x + 2y$

This is a linear function, and its domain is all real numbers for $x$ and $y$, i.e., $\mathbb{R}^2$.

Step4: Find domain of $f(x,y)=e^{\frac{1}{x + 2y}}$

The exponent $\frac{1}{x + 2y}$ is well - defined when $x+2y
eq0$, which means all of $\mathbb{R}^2$ except the line $y =-\frac{1}{2}x$.

1. $f(x,y)=\sqrt{x^{3}y^{5}}$ matches B (union of first and third quadrants).
2. $f(x,y)=\ln(x + 2y)$ matches A (region above $y =-\frac{1}{2}x$).
3. $f(x,y)=x + 2y$ matches C (all of $\mathbb{R}^2$).
4. $f(x,y)=e^{\frac{1}{x + 2y}}$ matches the region that is $\mathbb{R}^2$ minus the line $y=-\frac{1}{2}x$ (not shown in the given options in a detailed way, but based on the previous matches, the remaining logic).

Answer:

1. B
2. A
3. C