Suppose that the function $h:\mathbb{R}\to\mathbb{R}$ is differentiable and that there is a positive number $c$ such that $h'(t)\geq c$, for all points $t\in\mathbb{R}$. Prove that there is exactly one number $t$ at which $h(t)=0$.
Proof (BWOC)
Suppose that there are exactly $n$ numbers, $\{x_1,x_2,\dots,x_n\}\in\mathbb{R}$, $x_1<x_2<\dots<x_n$, for which $h(x_1) = h(x_2) = \cdots = h(x_n) = 0$.
Because $h$ is continuous, $h$ is either positive or negative for all $t_1\in (x_1,x_2)$, all $t_2\in(x_2,x_3)$, $\cdots$, all $t_{n-1}\in(x_{n-1},x_n)$.
Additionally, because $h'(t)$ is defined for all $t\in\mathbb{R}$, if $h$ is positive in $(-\infty, x_1)$, $h$ must become negative in $(x_1,x_2)$, else the derivative will not be defined at $x_1$. Likewise, $h$ must become negative in $(x_2,x_3)$.
Because $h$ transitioned from positive to negative to positive, $h'$ transitioned from negative to positive respectively. By the Intermediate Value Theorem, $h'(t^*) = 0$ for some $t^*$ in $(x_1,x_2)$. There are $n-1$ such $t^*$'s.
This is a contradiction, because $h'(t)>0\forall t\in\mathbb{R}$.
Therefore $h$ attains exactly one zero, so that the number of points at which $h'(t) = 0$ is exactly $0$.
The intermediate value theorem for derivatives is a bit much. If $h(x_1) = h(x_2) = 0$, then by Rolle's theorem there exists $t \in (x_1,x_2)$ satisfying $h'(t) = 0$, contrary to hypothesis.