Definition of a strictly increasing function What is the definition of strictly increasing for a function? Why does $f' &gt;0$ imply $f$ is strictly increasing?
Much obliged if anyone can help me .
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 $x&lt;y$ in $X$. If the inequality is strict, i.e., $f(x)&lt;f(y)$ whenever $x&lt;y$ 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 "$f'&gt;0$ implies $f$ strictly increasing" is a consequence of the mean value theorem (essentially, if $x&lt;y$ 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 $0$).
Let $A$ be the open subset where $f$ is differentiable. We know that $f'(x)&gt;0$ 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 $f'(z)&gt;0$, it follows that $\dfrac {f(x_1)-f(x_2)} {x_1-x_2}$, so $f(x_1)-f(x_2)&gt;0$, i.e. $f(x_1)&lt;f(x_2)$
(because $x_1&lt;x_2$). Therefore $f$ is strictly increasing in $A$.