Confusion between the definition of increasing function and a theorem regarding it.

The definition of increasing function given in my school maths text book is

Let I be an open interval contained in the domain of a real valued function f. Then f is said to be increasing on I if $a<b\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}f(a)\le f(b)$ for all $a,b\in I$.

And a theorem given after it is

f is increasing in I if ${f}^{\prime}(x)>0\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}x\in I$.

Shouldn't it be ${f}^{\prime}(x)\ge 0$.

Is constant function an increasing function? Or a function like $f(x)={x}^{3}$ where ${f}^{\prime}(x)=0$ at some or all points which are increasing according to definition.

Edit: The definition of decreasing function given is

f is decreasing on I if $a<b\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}f(a)\ge f(b)$.

A constant function follow this definition too.

The definition of increasing function given in my school maths text book is

Let I be an open interval contained in the domain of a real valued function f. Then f is said to be increasing on I if $a<b\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}f(a)\le f(b)$ for all $a,b\in I$.

And a theorem given after it is

f is increasing in I if ${f}^{\prime}(x)>0\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}x\in I$.

Shouldn't it be ${f}^{\prime}(x)\ge 0$.

Is constant function an increasing function? Or a function like $f(x)={x}^{3}$ where ${f}^{\prime}(x)=0$ at some or all points which are increasing according to definition.

Edit: The definition of decreasing function given is

f is decreasing on I if $a<b\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}f(a)\ge f(b)$.

A constant function follow this definition too.