Let $f'(x_0)>0$ show that there is a $\delta>0$ so that $f(x)<f(x_0)<f(y)$ whenever $x_0-\delta<x<x_0<y<x_0+\delta.$

Full Problem Statement: Let $f'(x_0)>0$ for some point $x_0$ in the interior of the domain $D$ of $f$ show that there is a $\delta>0$ so that $f(x)<f(x_0)<f(y)$ whenever $x_0-\delta<x<x_0<y<x_0+\delta.$

My Attempt: Since $x_0$ is an interior point of $I$ there exists a ball completely contained inside the interval. Let radius of that ball be $r$ and so we have that $(x_0-r,x_0+r)\subseteq D.$ Then if $x_0-r<x<x_0<y<x_o+r$ and $f(x_0)\geq f(y)$ then we have that $$f'(x_0)=f_{+}'(x_0)=\lim_{y\to x_0}\frac{f(y)-f(x_0)}{y-x_0}\leq 0$$ which is a contradiction. Similarily we can show that $f(x)<f(x_0).$ I am not sure whether this argument is correct and so would appreciate any insight.

• – Paramanand Singh Dec 16 '17 at 5:39
• The conclusion of the result in question is also stated as "$f$ is strictly increasing at $x_{0}$". – Paramanand Singh Dec 16 '17 at 5:40

No, the $y$ for which $f(x_{0})\geq f(y)$ is a particular $y$, but the $y$ in the limit $\lim_{y\rightarrow x_{0}}$ are all sufficiently closed $y$ to $x_{0}$.
Actually you can prove that statement in direct way. Find some $\delta>0$ such that for all $x\in(x_{0}-\delta,x_{0}+\delta)-\{x_{0}\}$ \begin{align*} \left|\dfrac{f(x)-f(x_{0})}{x-x_{0}}-f'(x_{0})\right|<\dfrac{1}{2}f'(x_{0}), \end{align*} expand the inequality get \begin{align*} \dfrac{f(x_{0})-f(x)}{x_{0}-x}-f'(x_{0})>-\dfrac{1}{2}f'(x_{0}),~~~~x\in(x_{0}-\delta,x_{0}), \end{align*} so \begin{align*} f(x_{0})-f(x)>\dfrac{1}{2}f'(x_{0})(x_{0}-x)>0,~~~~x\in(x_{0}-\delta,x_{0}), \end{align*} andn similarly, \begin{align*} f(y)-f(x_{0})>\dfrac{1}{2}f'(x_{0})(y-x_{0})>0,~~~~y\in(x_{0},x_{0}+\delta). \end{align*}