Problem regarding odd and even functions.

Question

Let $f$ and $g$ be two distinct functions defined on the set of real numbers such that $f$ is an odd function and $g$ is an even function. It is given that $$f'(x) > g'(x) \quad \forall \quad x \in \mathbf{R}$$

What can we say about the function $g$? Also, how many solutions of $x$ are there for the equation $f(x)=g(x)$?

Answer 1

Since $f'(x)>g'(x)$ You already know that the function $h(x)=f(x)-g(x) $ is strictly increasing function ( as $f'(x)-g'(x)>0$ ) thus $h(x)=0$ has at most one solution.

Thus $g(x)=f(x) $ has at most $1$ real solution.

Answer 2

(I'm assuming $f$ and $g$ are differentiable, in order for your equation to make sense).

For the first question, you can't say anything about $g$ beyond the assumption that it is even. Indeed for any even differentiable function $g$, let $$ f(x)=\int_0^x|g'(t)|+1\;dt $$ Since $g$ is even, $g'$ is odd, so $|g'(t)|+1$ is even, making $f$ odd. Clearly $$ f'(x)=|g'(x)|+1>g'(x). $$ For the second question, there is at most $1$ solution since $f(x)-g(x)$ is strictly increasing. There could be one (eg $f(x)=x$, $g(x)=0$) or zero (eg $f(x)=\arctan(x)$, $g(x)=-2$).

Answer 3

You can use graphical analysis. Since we know that $g(x)$ is an even function its graph is symmetrical about y axis, so that g(-x)=g(x).

Now, since f(x) is an odd function, its graph must necessarily pass through the origin.It is like this

Since it is given that $f'(x)\gt g'(x)$, both the graphs must intersects at atleast one point. Which in turn implies that $f(x)=g(x)$ must have atleast one solution.Looking like this-