# Problem of the Week Problem E and Solution Everything in its Place 3

## Problem

1. A Venn diagram has two circles, labelled A and B.

Each circle contains functions, $$f(x)$$, that satisfy the following criteria.

• A: $$f(2)=-3$$

• B: $$f(-2)=-1$$

The overlapping region in the middle contains functions that are in both A and B, and the region outside both circles contains functions that are neither in A nor B.

In total this Venn diagram has four regions. Place functions in as many of the regions as you can. Is it possible to find a function for each region?

2. A Venn diagram has three circles, labelled A, B, and C.

Each circle contains ordered pairs, $$(x,y)$$, where $$x$$ and $$y$$ are real numbers, that satisfy the following criteria.

• A: $$y=(x+3)^3+2$$

• B: $$y=\dfrac{1}{2}x^2+1$$

• C: $$y=|x+1|$$

In total this Venn diagram has eight regions. Place ordered pairs in as many of the regions as you can. Is it possible to find an ordered pair for each region?

## Solution

1. We have marked the four regions W, X, Y, and Z.

When creating functions, you can think of the problem algebraically, or graphically. When thinking algebraically, a function that satisfies $$f(2)=-3$$ is one that evaluates to $$-3$$ when $$2$$ is substituted for $$x$$. When thinking graphically, a function that satisfies $$f(2)=-3$$ is one whose graph goes through the point $$(2,-3)$$.

• Any function in region W must satisfy $$f(2)=-3$$ but not $$f(-2)=-1$$. There are infinitely many possibilities. Some examples are $$f(x)=-x-1$$ and $$f(x)=x^2-7$$.

• Any function in region X must satisfy both $$f(2)=-3$$ and $$f(-2)=-1$$. In other words, the graph of the function must pass through both $$(2,-3)$$ and $$(-2,-1)$$. There are infinitely many possibilities. Some examples are $$f(x)=-\frac{1}{2}x-2$$ and $$f(x)=-\frac{1}{8}x^3-2$$.

• Any function in region Y must satisfy $$f(-2)=-1$$ but not $$f(2)=-3$$. There are infinitely many possibilities. Some examples are $$f(x)=\frac{1}{2}x$$ and $$f(x)=x^2-5$$.

• Any function in region Z must satisfy neither $$f(2)=-3$$ nor $$f(-2)=-1$$. There are again infinitely many possibilities. Some examples are $$f(x)=x$$ and $$f(x)=x^2$$.

2. We have marked the eight regions S, T, U, V, W, X, Y, and Z.

We will name the functions as follows: $$f(x)=(x+3)^3+2$$, $$g(x)=\frac{1}{2}x^2+1$$, and $$h(x)=|x+1|$$. We have also provided a graph of the functions.

Note that the graphs of $$f$$ and $$g$$ intersect at $$(-2,3)$$, the graphs of $$f$$ and $$h$$ intersect at $$(-3,2)$$, and the graphs of $$g$$ and $$h$$ intersect at $$(0,1)$$ and $$(2,3)$$.

• Any ordered pair in region S must satisfy the equation $$y=f(x)$$, but not $$y=g(x)$$ or $$y=h(x)$$. Since the point $$(-4,1)$$ is on the graph of $$f$$, but not on the graph of $$g$$ or $$h$$, the ordered pair $$(-4,1)$$ works. There are infinitely many others as well.

• Any ordered pair in region T must satisfy the equations $$y=f(x)$$ and $$y=g(x)$$, but not $$y=h(x)$$. Since the point $$(-2,3)$$ is on the graph of $$f$$ and on the graph of $$g$$, but not on the graph of $$h$$, the ordered pair $$(-2,3)$$ works. In fact, since $$(-2,3)$$ is the only point of intersection of the graphs of $$f$$ and $$g$$, this is the only possible choice for region T.

• Any ordered pair in region U must satisfy the equation $$y=g(x)$$, but not $$y=f(x)$$ or $$y=h(x)$$. The ordered pair $$(4,9)$$ works. There are infinitely many others as well.

• Any ordered pair in region V must satisfy the equations $$y=f(x)$$ and $$y=h(x)$$, but not $$y=g(x)$$. The ordered pair $$(-3,2)$$ works. In fact, since $$(-3,2)$$ is the only point of intersection of the graphs of $$f$$ and $$h$$, this is the only possible choice for region V.

• Any ordered pair in region W must satisfy the equations $$y=f(x)$$, $$y=g(x)$$, and $$y=h(x)$$. There are no ordered pairs that satisfy this as the graphs of the three functions do not have a common point of intersection. So this region must remain empty.

• Any ordered pair in region X must satisfy the equations $$y=g(x)$$ and $$y=h(x)$$, but not $$y=f(x)$$. The ordered pairs $$(0,1)$$ and $$(2,3)$$ work. In fact, since these are the only points of intersection of the graphs of $$g$$ and $$h$$, these are the only possible choices for region X.

• Any ordered pair in region Y must satisfy the equation $$y=h(x)$$, but not $$y=f(x)$$ or $$y=g(x)$$. The ordered pair $$(-2,1)$$ works. There are infinitely many others as well.

• Any ordered pair in region Z must not satisfy the equations $$y=f(x)$$, $$y=g(x)$$, or $$y=h(x)$$. The ordered pair $$(0,0)$$ works. There are infinitely many others as well.