Let `f: R->R` be a continuous onto function satisfying `f(x)+f(-x)=0AAx in Rdot` If `f(-3)=2a n df(5)=4in[-5,5],` then the minimum number of roots of the equation `f(x)=0` is

Answer

3

Explanation

`f(x)+f(-x)=0`
Therefore,`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.