Let [math]f(x) = (x if x < 0) & (x + 1 if x \geq 0)[/math].

Let [math]f(x) = x if x < 0[/math] and [math]x + 1 if x \geq 0[/math].
The function is not continuous (because in [math]x = 0[/math] we have a jump discontinuity), but is the function still differentiable?
For [math]x <0 [/math] we have [math]f'(x) = 1[/math] and for [math]x > 0[/math] we also have [math]f'(x) = 1[/math].
Furthermore we have [math]\lim_{h\to 0^-} \frac{f(0 + h) - f(0)}{h} = \lim_{h\to 0^-} \frac{h}{h} = 1[/math] and [math]\lim_{h\to 0^+} \frac{f(0 + h) - f(0)}{h} = \lim_{h\to 0^+} \frac{h + 1 - 1}{h} = 1[/math], so we have [math]f_{-}^{'} (0) = f_{+}^{'} (0) = 1[/math], so isn't it differentiable [math]\forall x \in \R[/math]?
But every function that is differentiable is also continuous, isn't it? Where did I do wrong?
Thanks

[math] \begin{cases}
x & x < 0 \\
x + 1 & x\geq 0
\end{cases}
[\math]

Fuck latex man

I'll write it plaintext:
f(x) = x if x < 0
x + 1 if x >= 0

>>8490425
Don't blame Latex for your stupidity

>>8490413
>but is the function still differentiable?

You need to remember that differentiability is defined pontwise.

Given a function f, it is differentiable at a point x if the defined limit exists.

Then it is differentiable in an interval if it is differentiable in every point in that interval.

So obviously that function is not differentiable from (-inf,inf) because it is not differentiable at 0 but it is differentiable in many intervals. Furthermore, the piecing together of that broken derivative can be said to be its derivative.

That function looks looks like x and x+1 and the derivative for both is f(x) = 1 but only when x is not equal to 0, as this derivative also has a discontinuity on 0.

>>8490427
You can preview the tex by clicking the "TEX" in the upper left corner.

Differentiability implies continuity.

>>8490420
For h -> 0- you have f(0+h)-f(0)=h-1 since 0+h<0.

>>8490433
>>8490444
I knew there was a latex preview button but i could't find it...
>>8490436
So you're saying that I need to calculate the derivative of x and (x + 1) separetely, but because I restricted domain the derivative calculated in x = 0 of both of them doesn't exist. This implies that also the piecewise function can't be derived in x = 0. Did I understand it correctly? Thanks again

>>8490465
Check
>>8490454
you did the h->0- wrong.

>>8490425
[math]
\begin{cases}
x & x < 0 \\
x + 1 & x\geq 0
\end{cases}
[/math]

>>8490454
>>8490469
Maybe he's right: [math] lim_{h \to 0^-} \frac{f(x + h) - f(x)}{h}[/math].
For [math]x = 0[/math] we have [math]x + h < 0[/math], so we use [math]x[/math], but [math]f(x) = f(0)[/math], so we use [math]x + 1[/math].
In the end for [math]x = 0[/math] we get [math] lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} = lim_{h \to 0^-} \frac{h - (0 + 1)} {h} = lim_{h \to 0^-} \frac{-1} {h} = + \infty [/math].

Ok got it. Thanks!