Find the points at which the function is not differentiable

In summary, the function is not differentiable at x=0, and for x <= -√(3/2) and x >= √(3/2) the function can be written as f(x) = \left\{\begin{array}{l}\sqrt{1+x^2}~\text{if}~-2\leq x\leq -\sqrt{3/2}\\\sqrt{4-x^2}~\text{if}~-\sqrt{3/2}\leq x\leq 0\\\sqrt{1+x^2}~\text{if}~0
  • #1
zorro
1,384
0

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:
Physics news on Phys.org
  • #2


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

[tex]\sqrt{4-x}\leq \sqrt{1+x^2}[/tex]

and for what x that doesn't hold. This will make it easier for later...
 
  • #3


Is that a necessary step for such max-min kind of problems?
 
  • #4


You want to know what the function looks like, don't you? Then this is a necessary step...
 
  • #5


OK.
I found that out.
For x <= -√(3/2) and x >= √(3/2)
What next?
 
  • #6


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

First you need to calculate the equality

[tex]\sqrt{4-x}=\sqrt{1+x^2}[/tex]

This corresponds to the quadratic equation

[tex]x^2+x-3=0[/tex]

And when I solve this, I don't get what you get...
 
  • #7


Sorry. There is a mistake in my original post. It is (4 - x^2)^0.5.
Square is missing.
 
  • #8


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

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

Now, you can see that there are but a few points in which f is possible not differentiable...
 
  • #9


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

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

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:
 

1. What is the definition of differentiability?

The definition of differentiability is the ability of a function to be smoothly and continuously differentiable at a point, meaning that it has a well-defined slope or derivative at that point.

2. How can I determine if a function is differentiable at a given point?

To determine if a function is differentiable at a given point, you can use the definition of differentiability to calculate the derivative at that point. If the derivative exists and is continuous, then the function is differentiable at that point.

3. What does it mean for a function to be not differentiable?

If a function is not differentiable at a point, it means that the derivative of the function does not exist or is not continuous at that point. This can occur when there is a sharp corner, vertical tangent, or discontinuity at that point.

4. How can I find the points at which a function is not differentiable?

To find the points at which a function is not differentiable, you can look for any discontinuities or points where the derivative does not exist. You can also graph the function and visually identify any sharp corners or vertical tangents.

5. Why is it important to identify points of non-differentiability in a function?

Identifying points of non-differentiability in a function is important because it can help us understand the behavior of the function and any potential limitations or errors in our calculations. It also allows us to make appropriate adjustments to the function or use alternative methods to analyze it.

Similar threads

  • Calculus and Beyond Homework Help
Replies
26
Views
821
  • Calculus and Beyond Homework Help
Replies
7
Views
133
  • Calculus and Beyond Homework Help
Replies
1
Views
170
  • Calculus and Beyond Homework Help
Replies
14
Views
921
  • Calculus and Beyond Homework Help
Replies
3
Views
862
  • Calculus and Beyond Homework Help
Replies
10
Views
868
  • Calculus and Beyond Homework Help
Replies
2
Views
221
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
132
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
Back
Top