Let f : R→R be a continuous onto function satisfying f(x)+f(−x)=0∀x∈R. If f(−3)=2 and f(5)=4 in [−5, 5], then the minimum number of roots of the equation f(x) = 0 is _______.

f(x)+f(−x)=0Therefore, f(x) is an odd function.Since points (-3, 2) and (5, 4) lie on the curve, (3,-2) and (-5, -4) will also lie on the curve.For minimum number of roots, graph of continuous function f(x) is as follows:From the above graph of f(x), it is clear that equation f(x) = 0 has at least three real roots.