# Determine when f is differentiable

1. Dec 9, 2005

### SomeRandomGuy

How do you determine when f is differentiably from a real analysis standpoint (no graphs and calculus)? Would I simply look for a point of discontinuity? We have 4 problems on our homework assignment involving this issue and I don't see one example in my notes or the book adressing it. Here is one of our problems incase your wondering:

Determine the values of x for which f(x) is differentiable and find it's derivative:
f(x) = x|x|

Thanks

2. Dec 9, 2005

### JasonRox

Start by looking at f(x) = |x|, and using the definition of the derivative as x approaches 0.

Are the limits the same?

3. Dec 9, 2005

### SomeRandomGuy

I don't understand. Are which two limits the same? The definition of the derivative on |x| and x|x|?

4. Dec 9, 2005

### D H

Staff Emeritus
You misunderstood JasonRox' post. I'll expand on it. First, forget about $x{\lvert x\rvert}$ for a moment. Just look at ${\lvert x\rvert}$. When you are looking at that function, examine its behavior near zero. In particular, look at
$$\lim_{\Delta x\to 0}\frac{{\lvert x + \Delta x\rvert} - {\lvert x\rvert}}{\Delta x}$$
at $x=0$.

By two limits, JasonRox meant the limit as $\Delta x$ approaches zero from below (i.e., make $\Delta x$ some small negative number that approaches zero) versus above (same thing, but now make $\Delta x$ positive).

Last edited: Dec 9, 2005
5. Dec 9, 2005

### JasonRox

Thanks, for clearing it up.

Yeah, once you see the idea behind f(x) = |x|, you'll totally understand it for f(x) = x|x|.

Draw the graph as well, so you see what's happening.

6. Dec 9, 2005

### matt grime

of course if you remember |x|=x for x=>0 and -x otherwise then it is clear that the functions is differentialbe away from the origin.

7. Dec 9, 2005

### HallsofIvy

First of all, NO, it is NOT the case that "if a function is continuous at x= a, it is differentiable at x= a"! For example, f(x)= |x|. That is continuous at x= 0 but not differentiable at x= 0.

f(x)= x|x|= x2 if x> 0 which is obviously differentiable for x> 0.
f(x)= x|x|= -x2 if x< 0 which is obviously differentiable for x< 0.

$$lim_{h\rightarrow 0^+}\frac{f(h)- f(0)}{h}$$
and
$$lim_{h\rightarrow 0^-}\frac{f(h)- f(0)}{h}$$
?
Take the limit

8. Dec 9, 2005

### HallsofIvy

First of all, NO, it is NOT the case that "if a function is continuous at x= a, it is differentiable at x= a"! For example, f(x)= |x|. That is continuous at x= 0 but not differentiable at x= 0.

f(x)= x|x|= x2 if x> 0 which is obviously differentiable for x> 0.
f(x)= x|x|= -x2 if x< 0 which is obviously differentiable for x< 0.

$$lim_{h\rightarrow 0^+}\frac{f(h)- f(0)}{h}$$
and
$$lim_{h\rightarrow 0^-}\frac{f(h)- f(0)}{h}$$
?
Take the limit

9. Dec 9, 2005

### JasonRox

Yeah, totally right.

It is true that if it is differentiable at x=a, then it is continuous at x=a.

The converse is not true, just like HallsofIvy showed.

10. Dec 10, 2005

### SomeRandomGuy

Ok, I did what you guys said and took the right and left hand derivatives for this function.
I got the right hand derivative to be infiinity, while the left is -infinity. So, it's not differentiable at x = 0. I understand this, and I used this technique on the 4 other problems. I have one question, however.

For the function f(x)=|x^2-1|, I found it to be differentiable at x = -1, but not at x = 1. Intuitively, this is wrong, but I can't find my mistake. The right and left hand derivatives I got to equal 0. Is this correct?

11. Dec 10, 2005

### matt grime

12. Dec 10, 2005

### SomeRandomGuy

Thanks for the help guys, I made some pretty dumb mistakes in a few of my posts that I caught tonight after looking things over. I think I have the answers now. when f(x) = |x^2-1|, it's not differentiable at x = 1, -1. when f(x) = x|x|, it's differentiable at 0. Is this right? If needed, ill post work so you guys just don't think your giving me an answer.

13. Dec 10, 2005

### JasonRox

Just plot that the graph.

That's the easiest way to know.