Find the points at which the function is not differentiable

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Homework Statement


Find the points at which the function is not differentiable.

space;0&space;\\\\&space;=&space;min(\sqrt{4-x^{2}},\sqrt{1&plus;x^{2}})&space;\\0<x\leq&space;2.gif

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
 
Last edited:
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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...
 


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


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


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


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...
 


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


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&lt;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...
 


micromass said:
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&lt;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?
 
  • #10


Ah, yes, you are right. There should be strict inequalities instead of just an inequality. Sorry for the confusion...
 
  • #11


Ah, ok, I got it :biggrin:
 
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