calculus
Consider the following function $f : \Bbb R \to \Bbb R$:
$$f(x) = \begin{cases} \sin x & x \ge 0 \\ x^2 + e^x & x < 0 \end{cases}$$
I was wondering if the above piece-wise function was differentiable over the set of Real numbers R, and how to figure it out.
I know that differentiability implies continuity, so maybe I could check if it was continuous as a starter?
The moral is, a discontinuous periodic function can have continuous derivative. (More precisely, a periodic function whose left- and right-hand limits are unequal at some points can, nonetheless, have derivative with removable discontinuities.)
The trig functions and the principal branches of their inverses are good at creating this type of behavior. While it's true that $$ \tan(\arctan x) = x\quad\text{for all real $x$,} $$ it's not true that $$ \arctan(\tan x) = x\quad\text{for all real $x$,} $$ only that $$ \arctan(\tan x) = x\quad\text{for all real $x$ with $|x| < \pi/2$.} $$
Edit: The arctangent function (also known as "the principal branch of arctan"), shown bold in the graph below, is single-valued. The tangent function, however, is periodic with period $\pi$, so there are infinitely many branches of inverse tangent, any two differing by an integer multiple of $\pi$. For every integer $n$, we have $$ x = \arctan(\tan x) + n\pi\quad\text{for $(n - \tfrac{1}{2})\pi < x < (n + \tfrac{1}{2})\pi$.} $$
Consider the functions
$$
f(x) = \frac{1}{\sqrt{2}} \arctan(\sqrt{2} \tan x),\qquad
F(x) = \int_{0}^{x} \frac{dt}{1 + \sin^{2} t}.
$$
These functions have the same derivative $g$ for all real $x$ unless $x = (n + \frac{1}{2})\pi$ for some integer $n$; for these $x$, the value $f(x)$ is undefined, because $\tan(n + \frac{1}{2})\pi$ is undefined. Call these points poles of the tangent function.
As $x$ runs over the period interval $(n - \frac{1}{2})\pi < x < (n + \frac{1}{2})\pi$ between two poles of the tangent function, $\sqrt{2}\tan x$ runs once over the entire real line, so $\arctan(\sqrt{2}\tan x)$ runs once over the interval $(-\frac{1}{2}\pi, \frac{1}{2}\pi)$. The graph of $f$ consists of "repeated units", with a "jump" at each point $(n + \frac{1}{2})\pi$; the function $f$ "forgets" how many poles of tan lie between $0$ and $x$.
By contrast, the graph of $F$ "remembers" how many poles lie between $0$ and $x$. When $x$ crosses a pole of tan, $\sqrt{2}\tan x$ "jumps" from being "near $+\infty$" to being "near $-\infty$". Because area under the graph of $g$ accumulates continuously, the value of $F$ "continues" using the "next-highest" branch of arctan. Loosely, instead of "returning to $-\infty$ and re-using the same branch of arctan" (as $f$ does), $F$ continues by "following a dashed line from right to left" and starting on the next branch of arctan.
That is how $f$ and $F$ manage to have the same derivative (except at the poles of tan, where $f$ is undefined), but $f$ is periodic while $F$ extends continuously over the poles.
In the extended real numbers $[-\infty, \infty]$, in fact, arctan has a continuous extension, defined by $$ \arctan(\pm\infty) = \lim_{x \to \pm\infty} \arctan x = \pm\tfrac{1}{2}\pi. $$ This isn't much help with $f$, because now $f$ gets two values at each pole of tan, namely $\pm \frac{\pi}{2\sqrt{2}}$, but it does dictate why each time $x$ crosses a pole of tan, $F$ "climbs" to the next branch of arctan, and it shows why $F$ is continuous at the poles of tan.
No matter what value you prescribe to $f(0)$, the function you have defined is still bounded on each closed interval and discontinuous only at the point 0, hence integrable by your definition. In fact more is true: The integral of your function is equal to the integral of $2x-3$ on any closed interval.
Best Answer
That is a good idea. Now check the $x=0$ point. For the right side of the graph, $\sin 0 =0$. For the left side, $0^2+e^0=1$. It is not continuous at $x=0$, and is therefore not differentiable at $x=0$.