Find the points at which the function is not differentiable

[zorro]

Homework Statement

Find the points at which the function is not differentiable.

Homework Equations

It is not asked to check differentiability at a particular point.
How do I find the points which are not differentiable?
The function is not continuous at x=0

 

[micromass]

You'll first need to investigate your function a little bit.
You will first need to know for what x

$$\sqrt{4-x}\leq \sqrt{1+x^2}$$

and for what x that doesn't hold. This will make it easier for later...

 

[zorro]

Is that a necessary step for such max-min kind of problems?

 

[micromass]

You want to know what the function looks like, don't you? Then this is a necessary step...

 

[zorro]

OK.
I found that out.
For x <= -√(3/2) and x >= √(3/2)
What next?

 

[micromass]

Are you sure about that? I seem to be getting something else...

First you need to calculate the equality

$$\sqrt{4-x}=\sqrt{1+x^2}$$

This corresponds to the quadratic equation

$$x^2+x-3=0$$

And when I solve this, I don't get what you get...

 

[zorro]

Sorry. There is a mistake in my original post. It is (4 - x^2)^0.5.
Square is missing.

 

[micromass]

Ah, yes. Then you are correct.
So your function can actually be written as

$$f(x)=\left\{\begin{array}{l}<br /> \sqrt{1+x^2}~\text{if}~-2\leq x\leq -\sqrt{3/2}\\<br /> \sqrt{4-x^2}~\text{if}~-\sqrt{3/2}\leq x\leq 0\\<br /> \sqrt{1+x^2}~\text{if}~0<x\leq \sqrt{3/2}\\<br /> \sqrt{4-x^2}~\text{if}~\sqrt{3/2}\leq x\leq 2<br /> \end{array}\right.$$

Now, you can see that there are but a few points in which f is possible not differentiable...

 

[zorro]

Ah, yes. Then you are correct.
So your function can actually be written as

$$f(x)=\left\{\begin{array}{l}<br /> \sqrt{1+x^2}~\text{if}~-2\leq x\leq -\sqrt{3/2}\\<br /> \sqrt{4-x^2}~\text{if}~-\sqrt{3/2}\leq x\leq 0\\<br /> \sqrt{1+x^2}~\text{if}~0<x\leq \sqrt{3/2}\\<br /> \sqrt{4-x^2}~\text{if}~\sqrt{3/2}\leq x\leq 2<br /> \end{array}\right.$$

Now, you can see that there are but a few points in which f is possible not differentiable...

Hey I did not understand one thing.
You used equality sign between x and -√(3/2) in first two definitions.
How can a function be defined in 2 ways for the same domain? Am I wrong or was it your printing error?

 

[micromass]

Ah, yes, you are right. There should be strict inequalities instead of just an inequality. Sorry for the confusion...

 

[zorro]

Ah, ok, I got it