Answer
$f(-c)=-f(c)$, and there is a local maximum at $x=-c$.
Work Step by Step
Because f(x) is even, its graph is symmetric about the origin.
First, we know that $f(-c)=-f(c).$
Next, the part of the graph around $P=(c,f(c)$ is such that $f(c)$ is the smallest value of $f(x)$ for some interval around x=c.
There exists a positive $h$ such that $ x\in(c-h,c+h)\Rightarrow f(x)\geq f(c)$
Then, reflecting that part of the graph over the origin,
point P will be reflected to point $Q=(-c, -f(c))$, and the points that were OVER $f(c)$ will be reflected to points that are UNDER $-f(c)$.
This means that Q is a point of local maximum.
