What is the definition of strictly increasing for a function? Why does $f' >0$ imply $f$ is strictly increasing?
Much obliged if anyone can help me .
$\endgroup$4 Answers
$\begingroup$A function $f:X\to \mathbb R$ defined on a set $X\subset \mathbb R$ is said to be increasing if $f(x)\leq f(y)$ whenever $x<y$ in $X$. If the inequality is strict, i.e., $f(x)<f(y)$ whenever $x<y$ in $X$, then $f$ is said to be strictly increasing.
A similar definition for decreasing and strictly decreasing applies (simply reverse all inequalities).
If $X$ is connected (so, an interval), then the result "$f'>0$ implies $f$ strictly increasing" is a consequence of the mean value theorem (essentially, if $x<y$ and $f(x)=f(y)$, then there must exist a point $z$ with $x\leq z\leq y$ with $f'(z)=0$ since the slope of the secant between $(x,f(x))$ and $(y,f(y))$ would be $0$).
$\endgroup$ 3 $\begingroup$Strictly increasing means if $x_1 < x_2$ then $f(x_1) < f(x_2)$.
One way to define the derivative is
$f'(x_2) = \lim_{x_1 \rightarrow x_2} \frac{f(x_2)-f(x_1)}{x_2-x_1} > 0$
From $\frac{f(x_2)-f(x_1)}{x_2-x_1} > 0$ you can see, that if $x_2-x_1>0$ then so must be $f(x_2)-f(x_1)>0$ hence $f(x_2)>f(x_1)$.
$\endgroup$ 3 $\begingroup$$\endgroup$ $\begingroup$A function $f(x)$ is said to be strictly increasing on an interval $I$ if $f(b)>f(a)$ for all $b>a,$ where $a,b$ in $I.$
Conversely$,$ a function $f(x)$ decreases on an interval $I$ if $f(b)<=f(a)$ for all $b>a$ with $a,b$ in $I.$ If $f(b)<f(a)$ for all $b>a,$ the function is said to be strictly decreasing$.$
On the other hand, if $f(b)>=f(a)$ for all $b>a,$ the function is said to be (non-strictly) increasing.
Let $A$ be the open subset where $f$ is differentiable. We know that $f'(x)>0$ for every $x \in A$. $f$ satisfies the criteria of the mean value theorem in $[x_1,x_2] \subset A$, therefore there exists $z \in (x_1, x_2)$ so that $f'(z)= \dfrac {f(x_1)-f(x_2)} {x_1-x_2}$. Since $f'(z)>0$, it follows that $\dfrac {f(x_1)-f(x_2)} {x_1-x_2}$, so $f(x_1)-f(x_2)>0$, i.e. $f(x_1)<f(x_2)$ (because $x_1<x_2$). Therefore $f$ is strictly increasing in $A$.
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