Determine when f is differentiable

[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

 

[JasonRox]

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

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

Are the limits the same?

 

[SomeRandomGuy]

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

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

 

[D H]

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

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).

 

[JasonRox]

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).

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.

 

[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.

 

[HallsofIvy]

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

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.

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

 

[HallsofIvy]

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

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.

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

 

[JasonRox]

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.
Now what about
lim_{h\rightarrow 0^+}\frac{f(h)- f(0)}{h}
and
lim_{h\rightarrow 0^-}\frac{f(h)- f(0)}{h}
?
Take the limit

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.

 

[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?

 

[matt grime]

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

that seems remarkably unlikely

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?

obviosuly not: just look at the graphswhen x<-1, and x>1, the function is identically x^2-1,so the derivative is 2x then

when -1<x<1 the function is 1-x^2 so its derivative is -2x

what are the limits of these as x tends to plus and minus 1?

 

[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.

 

[JasonRox]

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.

Just plot that the graph.

That's the easiest way to know.