# Problem regarding mean value theorem

Let $f:\mathbb R\to \mathbb R$ be a continuous function, and let

• $S=$ set of all slopes of secant of the graph of the function $f$
• $T=$ set of all slopes of the tangent of the graph of the function

Question: (true or false with reason) If $S=T=\mathbb R$ where $\mathbb R$ is the set of all real number, then the function is differentiable everywhere.

I think the answer is false because though $S=T=\mathbb R$, there may exist a point at which $f$ is not differentiable. Please anyone give an example of this.

added:there may exists points on the graph which is sharp or has vertical tangent.here tangent and secant are geomertric objects and not trigonometric ratios.

• How do you define $T$ if $f$ is nothing differentiable ? – Gabriel Romon Dec 8 '16 at 10:52
• void set @LeGrandDODOM – Sathasivam K Dec 8 '16 at 10:54
• At some point $o \in \mathbb{R}$, construct a function such that $f$ has a vertical tangent and $f'(a < x < o) < 0$ and $f'(o<x<b) > 0$. Then just have $f'(c) = 0$ and $f'(d) = 0$ somewhere outside of $[a,b]$. Now $f'(x)$ covers $\mathbb{R}$, but has at least 1 point where it is not differentiable. An example: $$y = \left \{ \begin{array}{ll} \sqrt{|x^2 -x|} & x > 0 \\ \sqrt{|x^2+x|} & x \leq 0 \end{array} \right.$$ – kmeis Dec 8 '16 at 11:14
• There's a small typo in my function. It should be: $$y = \left\{ \begin{array}{ll} \sqrt{|x^2-x|} & x > 0 \\ -\sqrt{|x^2+x|} & x \leq 0 \end{array} \right.$$ See:desmos.com/calculator/ewvast8uum – kmeis Dec 8 '16 at 11:24
• For this particular case, I just looked at a function with a local extreme that was strictly positive and $g(0)=0$ to obtain the second condition. Then the first enforced by a $\sqrt{}$ and reflection. – kmeis Dec 8 '16 at 12:03

Consider a function $f$ such that for some point $o \in \mathbb{R}$ we let $f$ have a vertical tangent and $f'(a<x<o) < 0$ and $f'(0<x<b) > 0$. Then for two points outside of $[a,b]$ (such as $c<a$ and $b<d$) let $f'(c) = f'(d) = 0$.
$$y = \left \{ \begin{array}{ll} \sqrt{|x^2-x|} & x > 0 \\ -\sqrt{|x^2-x|} & x \leq 0 \end{array} \right.$$.