A difficult function question, seems unsolvable

Question

Let $f(x)$ be a function which satisfies $f(2014+x)=f(2014-x)$ for all values of $x$. If the graph of $y=f(x)$ has exactly 3 real rots, find the sum of these roots.

I have no idea how to begin. Any help?

Answer 1

Put $g(x)= f(2014+x)$ then we get $$g(-x) = g(x)$$

so if $x_0$ is a root for $g$ then is also $-x_0$ since $g$ is even. But if even function has 3 roots then one of them must be $0$. Can you continue?

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So if $x_1,x_2,x_3$ are roots for $f$ then are $x_1-2014$, $x_2-2014$ and $x_3-2014$ roots for $g$. Say $x_3-2014=0$ and $ x_1-2014=-(x_2-2014)$, so $$x_1+x_2+x_3 = 6042$$

Answer 2

$2014$ is an axis of mirror symmetry, so a root at $2014+x$ implies there is a root at $2014-x$.

Fix some root $x_1$. We have $f(x_1)=0$, and $x_1=2014+\lambda$ for some $\lambda$. We hence also have $f(2014+\lambda)=f(2014-\lambda)$. Assuming that $\lambda\neq0$, this tells us that $x_1=2014+\lambda$ and $x_1=2014-\lambda$ are two of our roots.

Now, we have one more root, $x_3$, which can be written $x_3=2014+\varphi$, where $\varphi \neq \lambda$. But this also tells us that $f(2014-\varphi)=0$, so $2014-\varphi$ is a root. If $\varphi \neq -\varphi$, this would mean that we actually have 4 distinct roots, so this cannot be. Therefore, $\varphi=0$.

So the sum of our roots is $$x_1+x_2+x_3=(2014+\lambda)+(2014-\lambda)+(2014+\underbrace{\varphi }_{=0})=3 \cdot 2014=6042.$$

Answer 3

Consider some function t(x) such that t(x) = f(x+2014). Then we see that t(x)=t(-x). This means that t is an even function. Let us call the roots of this function $r_1, r_2,$ and $r_3$, labeling them from left to right. Because t is even, f($r_1$) = f(-$r_1$)= f($r_2$)=f(-$r_2$)=f($r_3$)=f(-$r_3$). But here we have six roots. You said that it had 3. We must have over-counted. How could we have done that? One explanation is that -$r_1$=$r_3$ and vice versa. But this only brings us down to 4 roots. We need three. The solution, therefore, requires that $r_2$ = $-r_2$. There, we have three roots and all is as it should be. So what does this mean? Well, if x=-x, then x must be 0. Therefore $r_2$ is zero. This means that the root of f which corresponds to the $r_2$ of t is 2014. Similarly, the other roots are $2014+ r_1 $ and $2014 - r_1$. Add these up to get 2014 $\times$ 3, or 6042. Your question is not unsolvable. I hope that this helps!