Cusps, which are not differentiable, often exist within absolute value functions and at certain points of piecewise functions.

My question is... can piecewise functions be differentiable throughout? Well, you know, suppose two functions are connected to each other in a such perfect way that there's no cusp such as, for example, when the slope is exactly zero from (a,b) in which c, which is between (a,b) is the x-value of the connection between two completely different functions. This doesn't happen often; cusps usually are present. But there are exceptions. So can piecewise functions be differentiable throughout if there's no cusp?

2. A cusp is a point at which two branches of a curve meet such that the tangents of each branch are equal... so at a cusp you are not differentiable as the derivatives from both directions are unequal.

A piecewise function can be differentiable throughout though, for example

f(x)= x^2 for x<1
f(x)= x/2 otherwise

at the the point x=1 (where the pieces meet) both the derivative from the left and the derivative from the right are 1/2

At all other points, f(x) is differentiable since x^2 and x/2 are differentiable everywhere

3. Hello, Kaitosan!

Can piecewise functions be differentiable throughout?
They certainly can . . . We can easily create one.

The graph must be continuous throughout and have equal derivatives at the "junctions".

Consider the graph of the parabola $y \:=\:x^2$
Bisect it through its axis of symmetry.
Move each "half" one unit to the left and right, respectively.

The graph looks like this:
Code:
                    |
*             |             *
|
*            |            *
*           |           *
*         |         *
- - - - - - *-----+-----* - - - - - -
-1     |     1
And we have: . $f(x) \;=\;\begin{Bmatrix} -(x+1)^2 & & x \leq -1 \\ 0 & & -1 < x < 1 \\ (x\:{\color{red}-}\:1)^2 & & x \geq 1 \end{Bmatrix}$

This is the example I gave when asked this very question
. . on an exam while in college . . . centuries ago.

Yes . . . a silly typo . . . Thanks for the heads-up.

4. Originally Posted by Soroban
Hello, Kaitosan!

They certainly can . . . We can easily create one.

The graph must be continuous throughout and have equal derivatives at the "junctions".

Consider the graph of the parabola $y \:=\:x^2$
Bisect it through its axis of symmetry.
Move each "half" one unit to the left and right, respectively.

The graph looks like this:
Code:
                    |
*             |             *
|
*            |            *
*           |           *
*         |         *
- - - - - - *-----+-----* - - - - - -
-1     |     1
And we have: . $f(x) \;=\;\begin{Bmatrix} -(x+1)^2 & & x \leq -1 \\ 0 & & -1 < x < 1 \\ (x+1)^2 & & x \geq 1 \end{Bmatrix}$

This is the example I gave when asked this very question
. . on an exam while in college . . . centuries ago.
.
Didn't you mean $(x-1)^2$ instead of $-(x+1)^2$?

Also, isn't the domain supposed to be switched between the two functions?

Anyways, thanks for answering my question. So... as long as there's no cusp, a piecewise function is differentiable! Gotcha.

5. Originally Posted by Kaitosan
Didn't you mean $(x-1)^2$ instead of $-(x+1)^2$?
Also, isn't the domain supposed to be switched between the two functions?
You are correct. To match the graph, it should be:
$f(x) \;=\;\begin{Bmatrix} (x+1)^2 & & x \leq -1 \\ 0 & & -1 < x < 1 \\ (x-1)^2 & & x \geq 1 \end{Bmatrix}$