Roots of a function of only negative slope If $f(x)$ is a differentiable function with $f'(x)<0$ for all $x$, then the equation $f(x)=0$ has at most one solution.
I do not understand why that should be true, please help.
 A: Suppose the function had two roots, $a$ and $b$. Then $f(a)=f(b)=0$. Applying the mean value theorem, we get that there is some $c$ between $a$ and $b$ such that $$\frac{f(a)-f(b)}{a-b}=f'(c).$$ Why is this a problem?
A: My thinkink is this:


*

*When $f'(x) < 0$ for any $x$, then the function is a decreasing function because slope for any point is negative.

*Since slope is negative everywhere, the function is only decreasing because there is no $x$ in $f'(x)$ that would be $> 0$. 

*Since it is only decreasing, it only goes through the x-axis once and no more than once because in order to cross the x-axis more than one time, it would have to be increasing in some point below the x-axis, so that's why $f(x)=0$ has only one solution.
Let's for example take a look at this function: $f(x)=-x^3$. Now, $f'(x)=-3x^2$. If you plug in any real $x$ for $f'(x)$, it will always be less than $0$ unless, of course, $x=0$ which is the solution for $f(x)=0$. So, we know by this, that the function is decreasing. You can also graph it to see that it is only decreasing. It also has only one solution, as slope of every point is negative.
