# Functions with no derivatives

1. Apr 16, 2006

### PPonte

This exercise asks to determine the values of x for which the following function do not have derivative:

1. f(x) = |x - 1|
2. f(x) = |x2 - 2x|

I draw the functions and saw which were the points that had an derivative in the right different from the one in the left. For those were the points of the function that do not have derivative. But there isn't another way to solve it? I came across that those points were the roots of the functions.

2. Apr 16, 2006

### Astronuc

Staff Emeritus
The functions would not have derivatives at the roots, since the derivatives at x = r - $\epsilon$ and x = r + $\epsilon$ as $\epsilon$ -> 0 are not equal.

3. Apr 16, 2006

### Beam me down

The following works on 1. f(x) = |x - 1|:

Find the derivitive. Then you should get (x-1)/|x-1|, this will yield positive or negative one, for all values of x except one (and maybe infinity, but that is aunimportant).

However when you take the limit when x = 1 you get: $$\lim_{h\rightarrow 0} \frac{h}{\mid h \mid}$$ However you can take the limit from the left or right. But when you do this:
$$\lim_{h\rightarrow 0^-} \frac{h}{\mid h \mid} \neq \lim_{h\rightarrow 0^+} \frac{h}{\mid h \mid}$$

Therefore the derivitive of a function is not defined at a point if the limit at that point from the left does not equal the same thing when the limit is taken from the right.

4. Apr 16, 2006

### PPonte

Sorry, Astronuc. I'm actually starting the study of derivatives so I do not know what $\epsilon$ and r refer to. Thank you, anyway.

Beam me down, thank you. Good point!

5. Apr 16, 2006

### HallsofIvy

Don't you know the definition of "derivative"? I assume that's where you would "start". $\epsilon$ is in the definition of derivative and r is just the value of x at which you are taking the derivative.

Beam me Down: How do you get that as the derivative of |x-1|? How would you differentiate |x|?

6. Apr 16, 2006

### Beam me down

$$\lim_{h\rightarrow 0} \frac{\mid x -1 + h \mid - \mid x -1 \mid}{ h }$$

When x = 1:

$$\lim_{h\rightarrow 0} \frac{\mid 1 -1 + h \mid - \mid 1 -1 \mid}{ h }$$

$$\lim_{h\rightarrow 0} \frac{\mid h \mid}{ h }$$

7. Apr 17, 2006

### HallsofIvy

which is what, precisely?

8. Apr 18, 2006

### Beam me down

Which just happens to equal zero, and is completely useless.

9. Apr 18, 2006

### HallsofIvy

No, it isn't- that limit does not exist! However, what I originally asked was "How did you arrive at the fact that the derivative of |x-1| is $\frac{x-1}{|x-1|}$, not necessarily at x= 1 since you then use that general derivative to show that it does not exist at x= 1. Actually what you did was argue that since the limit of that derivative function does not exist at x= 1, the derivative must not exist which is not true- the derivative of a function is not necessarily continuous.