Find the points at which the function is not differentiable

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Homework Help Overview

The discussion revolves around identifying points at which a given function is not differentiable. The function involves square root expressions and is defined piecewise over different intervals.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the necessary steps to determine differentiability, including checking the continuity of the function and solving inequalities related to the function's definition.
  • Questions arise regarding the correctness of the function's piecewise definition and the implications of using equality versus strict inequalities in the intervals.

Discussion Status

The conversation is active, with participants exploring different interpretations of the function's definition and its implications for differentiability. Some guidance has been offered regarding the need to investigate specific conditions for differentiability.

Contextual Notes

There is mention of a mistake in the original function definition, which has led to some confusion about the intervals and the conditions under which the function is defined. Participants are also clarifying the nature of inequalities used in the piecewise definition.

zorro
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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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