2

I'm trying to understand the solution to this problem:https://i.stack.imgur.com/HOGDV.jpg. The solution states that the answer is $C$ and $D$, because if $-p(x)=p(y)$, then the equation of the values that satisfy the $-p(x)=p(y)$ must by symmetric with respect to$ y=-x$. I have 2 questions about this.

$1$. Why does the fact that $-p(x)=p(y)$ mean that the equation must be symmetrical over y=-x? Is this only true for polynomials, or is it true for any function?

$2$. Does the fact that $p(x)=p(y)$ mean that the equation must be symmetrical over y=x? Is this only true for polynomials, or is it true for any function?

Thanks in advance

Mathguy
  • 33
  • The image shows 4 plots. On each, one could find a value for x1 and another one for x2 where p(x1)+p(x2)=0. I am surprised the question asks for 1 choice only. – NoChance Aug 18 '19 at 12:42
  • @NoChance Oh no, the question lets you make to choices. The right answer's C and D. I just didn't screenshot the multiple choice part (sorry about that). – Mathguy Aug 18 '19 at 12:42
  • For $y=x$ we get $2p(x) =0$ , for every $x$ , so $p$ must be the zero polynomial. None of these graphs represent it. – dmtri Aug 18 '19 at 12:44

3 Answers3

0

Let $S$ be the set of points $(x, y)$ such that $p(x) + p(y) = 0$.

Suppose $(x, y) \in S$. Then $p(x) + p(y) = 0$. But then also $p(y) + p(x) = 0$, and so $(y, x) \in S$. This means that $S$ is symmetric with respect to the line $y = x$. Thus, the answer (A) is wrong.

In general, $S$ might not be symmetric with respect to the line $y = -x$. Indeed, if $p(x) = x + 1$, then the equation $p(x) + p(y) = 0$ amounts to $x + y + 2 = 0$, which represents a line parallel to the line $y = -x$.

Luca Bressan
  • 6,845
0

Let $p$ be a real-valued function defined on some nonempty subset $D$ of $\mathbb{R}$.

If $G$ is the graph of the equation $p(x)+p(y)=0$, then: \begin{align*} &\text{The point}\;(a,b)\;\text{is on}\;G\\[4pt] \iff\;&p(a)+p(b)=0\\[4pt] \iff\;&p(b)+p(a)=0\\[4pt] \iff\;&\text{The point}\;(b,a)\;\text{is on}\;G\\[4pt] \end{align*} hence $G$ is symmetric about the line $y=x$.

Similarly, if $H$ is the graph of the equation $p(x)-p(y)=0$, then: \begin{align*} &\text{The point}\;(a,b)\;\text{is on}\;H\\[4pt] \iff\;&p(a)-p(b)=0\\[4pt] \iff\;&p(b)-p(a)=0\\[4pt] \iff\;&\text{The point}\;(b,a)\;\text{is on}\;H\\[4pt] \end{align*} hence $H$ is also symmetric about the line $y=x$.

Moreover, for all $d\in D$, the point $(d,d)$ is on $H$.

However, neither of the graphs $G,H$ is necessarily symmetric about the line $y=-x$.

For example:

  • If $p(t)=t-1$, the graph $G$ of the equation $p(x)+p(y)=0$ is the graph of the equation $x+y=2$, and $G$ is not symmetric about the line $y=-x$.$\\[4pt]$
  • If $p(t)=t^2-2t$, the graph $H$ of the equation $p(x)-p(y)=0$ is the union of the graphs of the equations $x+y=2$ and $x=y$, and $H$ is not symmetric about the line $y=-x$.
quasi
  • 58,772
-1

$C$ is true because you can define $p(x)=x^2-\frac{1}{2}$ and you have that

$p(x)+p(y)=x^2+y^2-1=0$

That is the equation of the circumference;

$D$ is true because you can define

$p(x)=x$

and you have that

$p(x)+p(y)=x+y=0$

that is the equation of the second and fourth quadrant;

$A$ is not true because it is the locus of $y=ax^2$ where $a>0$

Federico Fallucca
  • 8,593
  • 1
  • 9
  • 20