Suppose $f,g$ are two odd functions whose derivatives are continuous  
I was thinking about the above problem. Let us assume that $ \phi(x)=f(x)-g(x)$.Now i have to find out the value of $ \phi(x)=0$ for all $x \in [-1,1].$ I see that $ \phi'(0)<0$.Also $f(-1)<0$ and $f(1)>0$ and so there exists a point $c \in [-1,1]$ such that $f(c)=0.$But i can not progress from here.
Can someone point me in the right direction?Thanks everyone in advance for your time.
 A: Your $\phi=f-g$ satisfies $\phi(0)=0$, $\phi'(0)<0$, $\phi(1)>0$, $\phi$ is odd, and $\phi$ is continuously differentiable.
Because $\phi(x)=\int_0^x \phi'(t)dt$, $\phi'(0)<0$, and $\phi'$ is continuous, it follows that $\phi(x)<0$ for sufficiently small positive $x$.  Because $\phi(1)>0$, this implies that $\phi$ has a positive zero by the Intermediate Value Theorem.  Because $\phi$ is odd, this implies that $\phi$ has a negative zero.  Thus, $\phi$ has at least $3$ zeros.
I suggest showing that exactly 3 is possible with an example, starting by drawing pictures of functions satisfying the conditions.
A: As Jonas points out, $f(0)=g(0)=0$. Thus, if we can show that there must be at least one more, all we need to do is show by example that there can be two such functions with $3$ roots. The fact that it has at least two roots is obvious since $f'(0)<g'(0)$. This says that there is a small interval around $0$ for which $f(x)<g(x)$. Since $g(x)<1$ for all $x$ and $f(1)=1$, by the intermediate value theorem, we know there must be at least one more point where they are equal.
Can you come up with an example for the two functions?
