Let f:R → R be a continuos onto function satisfying $f\left(x\right)+f(-x)=0, \forall x\in R$. If f(-3)=2 and f(5) =4 in [-5,5]. Then the minimum number of roots of the equation f(x) = 0 is 

$\text{f(x) + f(-x) = 0}$

$\begin{array}{l}\text{f(x) is odd functions}\\ \text{Since points (-3,2) and (5,4) lie on the curve}\\ \therefore \text{(3,-2) and (-5,-4) will also lie on the curve}\\ \text{For minimum number of roots, graph of continuos function f(x) is as follows}\end{array}$

$\text{From the above graph of f(x) , it is clear that equation f(x) = 0 has at least three real roots}$