 # 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 Rightarrow f(a) leq f(b) for all a, b in I. And a theorem given after it is f is increasing in I if f′(x)>0 forall x in I. Shouldn't it be f′(x) geq 0. crazygbyo 2022-08-12 Answered
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 for all $a,b\in I$.
And a theorem given after it is
f is increasing in I if ${f}^{\prime }\left(x\right)>0\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall }x\in I$.
Shouldn't it be ${f}^{\prime }\left(x\right)\ge 0$.
Is constant function an increasing function? Or a function like $f\left(x\right)={x}^{3}$ where ${f}^{\prime }\left(x\right)=0$ 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.
A constant function follow this definition too.
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Step 1
Some would call $\mathrm{\forall }x\in I,{f}^{\prime }\left(x\right)\ge 0$ a non decreasing function.
Also some would call $\mathrm{\forall }x\in I,{f}^{\prime }\left(x\right)>0$ a strictly increasing function.
But, yes, $\mathrm{\forall }x\in I,{f}^{\prime }\left(x\right)\ge 0$ is often defined as an increasing function. In that case, the function $f\left(x\right)=1$ is an increasing function.
Sometime the word monotone gets thrown in there too.
You may as well get used to the fact that not all mathematicians use the same definition for things. As long as the book is consistent, that is acceptable.
Step 2
INCREASING:
- $\mathrm{\forall }x,y\in I,x
- $\mathrm{\forall }x\in I,{f}^{\prime }\left(x\right)\ge 0$
STRICTLY INCREASING:
- $\mathrm{\forall }x,y\in I,x
- $\mathrm{\forall }x\in I,{f}^{\prime }\left(x\right)>0$
MOTONIC:
- Strictly increasing or strictly decreasing