14.2 Functions of two variables

Once you pick the number for the range, you're basically done....

Any constant function, for example $f(x,y)=0$.

Any constant function will do. For example, $f(x,y)=0$ or $f(x,y)=1$.

This is a review of high-school material, since we have functions of only one variable. We want you to think about it to get in the right mindset.

Remark: because we're going off imprecise sketches, it wouldn't be wrong to give open intervals, rather than closed intervals, as your answers.

If you set $x=y=1$, is there a solution to the equation?

yes

If $x=1=y$, and $(x,y,z)$ is a point on the function, then:

So yes, $(1,1)$ is in the domain.

There's some fine print here. There are two different values of $z$ corresponding to the input $(x,y)=(1,1)$. That means that globally, $z$ isn't a function of $x$ and $y$, because a function should only ever have at most one output for any one input. Implicitly-defined functions often have this characteristic: it's not possible to write $z=f(x,y)$ for any single function $f$ of $x$ and $y$.

To find the range, consider all points in the domain with $x=0$.

Domain: all of $\mathbb R^2$. Range: $[0,\infty)$

The only part of the function that could possibly limit the domain is the square root: we must not try to take the square root of a negative number.

The expression $4x^2+y^2$ gives nonnegative numbers for any real values of $x$ and $y$. So no matter what $(x,y)$ we input, there is no danger of taking the square root of a negative number. So, the domain is all of $\mathbb R^2$.

We've already noted that $4x^2+y^2$ will give us numbers from $[0,\infty)$, but we should check whether it gives us all of those numbers. Indeed, if we set $x=0$, we see

the range of which is $[0,\infty)$.

So by choosing $x=0$ and the appropriate $y$, we can indeed get $f(x,y)$ to be any nonnegative number we desire. So, the range of $f$ is $[0,\infty).$

For the range, consider $h(x,0)$.

Domain: all of $\mathbb R^2$. Range: $[0,\infty)$.

The only restriction on our domain is that we can't divide by 0, and $1+y^2$ is never 0. So, our domain is all of $\mathbb R^2$.

Since $x^2 \geq 0$ and $1+y^2\geq 0$, we see first that $h(x,y)$ is never negative. The question now is whether it can actually achieve all nonnegative real values. If we set $y=0$, then $h(x,0)=x^2$, which has range $[0,\infty)$. So we can indeed find a point $h(x,y)=h(x,0)$ equal to any nonnegative number our hearts desire. That is, the range of $h(x,y)$ is $[0,\infty)$.

The domain of the function $\arcsin (x)$ is $[-1,1]$, and its range is $\left[ -\frac{\pi}{2},\frac{\pi}{2}\right]$.

Domain: interior of the unit circle. Range: $[0,\pi/2]$.

Recall the domain of the function $\arcsin (x)$ is $[-1,1]$, and its range is $\left[ -\frac{\pi}{2},\frac{\pi}{2}\right]$.

Since we can only put numbers from $[-1,1]$ into arcsine, we require for our domain

The left part of the inequality isn't hard, since $x^2+y^2$ is never negative. The right side tells us

i.e. $(x,y)$ is inside (or on) the unit circle.

Subject to the constraint $x^2+y^2 \le 1$, the domain of $x^2+y^2$ is $[0,1]$. The range of $\arcsin x$ subject to the constraint $0 \le x \le 1$ is $\left[0,\frac{\pi}2\right]$.

Red dotted line: range of $x^2+y^2$ subject to restrictions.
Blue solid line: range of $\arcsin (x^2+y^2)$.

One way of thinking of $xy>0$ is that $x$ and $y$ must have the same sign (and both be nonzero).

Domain: all points $(x,y)$ such that $x$ and $y$ have the same sign; $x$ and $y$ are nonzero; and $y \neq \frac1x$.

Range: $(-\infty,0)\cup(0,\infty)$.

To find the domain of $g$, there are two potential limiting issues: we can't divide by 0, and we can't take the logarithm of a nonpositive number.

Combining these two restrictions, the domain of $g(x,y)$ is all points $(x,y)$ such that $x$ and $y$ have the same sign; they are nonzero; and $y \neq \frac1x$. These points are graphed below. Dashed lines indicate points that are not in the domain.

With these restrictions, $xy$ can be any nonnegative number except 1; which means $\log(xy)$ can be any real number except 0; and finally the range of the entire function is $(-\infty,0)\cup(0,\infty)$. (This is illustrated in graphs below.)

Red dotted line: values of $xy$. Blue dashed line: values of $\log(xy)$

Blue dashed line: values of $\log(xy)$. Green solid line: values of $\frac{1}{\log(xy)}$

$y$ doesn't impact the final value of $f(x,y)$, so think of this as a problem from last semester. What are the maximum and minimum values of the function $f(x)=\frac{x^2}{x^2+1}$? Can you sketch its graph?

Domain: all of $\mathbb R^2$. Range: $[0,1)$.

The only thing that might limit the domain of this function is dividing by zero; but since $x^2+1>0$ for all real values of $x$, we see the domain of $f$ is the entire plane $\mathbb R^2$.

Since $y$ doesn't impact the value of $f$, we can consider the single-variable function

Since $f(x,y)=g(x)$ for any $(x,y)$, the range of $g$ will be the same as the range of $f$. Note $g(x)$ is continuous over all real numbers. So, its range will be (global min)$\leq g(x) \leq$ (global max). To help picture how $g(x)$ behaves, note further that $z=g(x)$ has a horizontal asymptote at $z=1$, and $g(x)$ is an even function. Let's find the critical points of $g(x)$.

The only CP of this function is $x=0$. Its horizontal asymptotes are 1 in both directions. So, the basic shape of the function is:

So, its range is $[0,1)$.

Consider the functions $f_1(x)=\frac{x}{x^2+1}$ and $f_2(y)=\sin y$ separately.

Domain: all of $\mathbb R^2$. Range: $\left[-\frac32,\frac32\right]$.

The domain of $f(x,y)$ is all of $\mathbb R^2$: the only possible restriction is dividing by zero, but $x^2+1>0$ for all values of $x$.

We can write $f(x,y)$ as

where $f_1(x)=\frac{x}{x^2+1}$ and $f_2(y)=\sin y$. Since there is not term depending on both $x$ and $y$, the maximum value of $f$ will occur when $x$ maximizes $f_1$ and $y$ maximizes $f_2$. Similarly, the minimum value of $f$ will occur when $x$ minimizes $f_1$ and $y$ minimizes $f_2$. Since these two functions are both continuous, we see that the range of $f$ will be

The range of $f_2(y)=\sin y$ is easy: it's $[-1,1]$. Let's consider $f_1(x)=\frac{x}{x^2+1}$. Note its horizontal asymptotes are 0 in both directions, and it's an odd function. To find its extrema, let's sketch it, starting by finding its critical points.

The CPs of $f_1$ are $x=1$ and $x=-1$.

To sketch $f_1$, let's find the sign of its first derivative on the intervals between its critical points.

Now we have enough information to sketch $z=f_1(x):$

So, the range of $f_1(x)$ is $\left[-\frac12,\frac12\right]$.

All together, the range of $f(x,y)$ is $\left[-\frac32,\frac32\right]$.

Do you see any signs that might point you in the right direction?

For example: domain should be all $(a,p)$ where $a \ge 0$ and $p>0$; range should be $[0,\infty)$.

Some general assumptions might be that the amount of money spend on advertisements shoudn't be negative, so we should have $a \ge 0$. Similarly, it's reasonable to assume that the company is not giving away its product, nor paying people to take it, so $p>0$. Finally, people won't demand a negative number of goods, so the range should be nonnegative.

That is one way of thinking about the problem, but different models might have different restrictions. For example, from time to time (including a time in 2020) oil futures trade at negative values: people were paying to give them away. So for certain models, negative prices and negative demands do make sense.

For other models, also an upper bound of some sort probable makes sense. Maybe you aren't able to sell more than one million of your product, because you don't have the capacity to manufacture more. Maybe demand will never exceed one product per person in your area. Such restrictions would further impact the domain and range that make sense for your model.

The domain will look like a ring

$\frac15 \leq x^2+y^2 \leq \frac13$: that is, the points $(x,y)$ that are inside or on the circle centred at the origin with radius $\frac{1}{\sqrt 3}$, but not inside the circle centred at the origin with radius $\frac{1}{\sqrt 5}.$

For this question, we solve two inequalities.

So, the points $(x,y)$ must be both:

First work with the function

Then, think about the implications of $t=x^2-y$.

The point $(x,y)$ must be in one of the following regions:

The bracketing in the definition of $g(x,y)$ is suggestive. If we define $t=x^2-y$, then we get the function

This is easy enough to graph using tools from last semester.

Sketched below is $z=72t^2-t^4$, with parts in the model range highlighted.

To find the $t$-values that correspond to the model range, we solve:

Similarly,

So, now we can fill in our sketch with $t$-values:

So we need to have $t$ in $[-\sqrt{68},-\sqrt{47}]\cup [-5,-2]\cup[2,5]\cup[\sqrt{47},\sqrt{68}]$.

Now, recall we used $t=x^2-y$. So if we have $a \leq t \leq b$, then this gives us two inequalities:

and

So, $t$ in the interval $[a,b]$ implies that $(x,y)$ must satisfy $x^2-b \leq y \leq x^2-a$:

We have four such possible intervals. All together, the point $(x,y)$ must be in one of the following regions: