Is this function differentiable at -π/2? Is it differentiable at 0?

In summary: But you should try to do them yourself first.In summary, the function f(x) is a piecewise function with different rules for different intervals of x. The question is whether this function is differentiable at certain values of x. The general approach is to first determine if the function is continuous at the point of interest, and then apply the limit definition of the derivative. The function must be both continuous and have a limit of the derivative at the point in order to be differentiable there. To solve this problem, it may be helpful to plot the graph and evaluate the derivatives of the left and right side functions at the point in question. Online tools such as wolframalpha.com can also be used to check the derivatives.
  • #1
SixBooks
5
0
The function f: R → R is: f(x) =

(tan x) / (1 + ³√x) ; for x ≥ 0,

sin x ; for (-π/2) ≤ x < 0,

x + (π/2) ; for x < -π/2
_

For -π/2 I would say it is not differentiable since both π and 2 are constants and you can not vary a constant.

For 0, I would derivate it by using the first function - (tan x) / (1 + ³√x) - and just replace x for 0. It gives me 0 as result. If x ≠ 0, the result could also be ≠ 0.

From here I can't go further...
 
Physics news on Phys.org
  • #2
SixBooks said:
The function f: R → R is: f(x) =

(tan x) / (1 + ³√x) ; for x ≥ 0,

sin x ; for (-π/2) ≤ x < 0,

x + (π/2) ; for x < -π/2
_

Your thread title is confusing. In English, we ask, "Is this function differentiable at 0?", for example.

As for the content of your post, I believe you are being asked if the piecewise function f(x) has a derivative at certain values of x. What you are supposed to do is apply the definition of the derivative to this function at the indicated values of x and decide if the definition is satisfied. If the definition is satisfied for a certain value of x, then the function is differentiable at that point.

The general approach is to first establish if the function is continuous at the point of interest. If it is not, no further work is needed; a function must be continuous at a point for a derivative to exist. The second part is to apply the limit definition of the derivative at the point of interest, and see if the limit exists. If the function is continuous and the limit of the derivative exists at the point of interest, then the function is differentiable at that point.

http://en.wikipedia.org/wiki/Derivative
For -π/2 I would say it is not differentiable since both π and 2 are constants and you can not vary a constant.

For 0, I would derivate it by using the first function - (tan x) / (1 + ³√x) - and just replace x for 0. It gives me 0 as result. If x ≠ 0, the result could also be ≠ 0.

From here I can't go further...

This argument makes no sense. Isn't 0 a constant, like both π and 2? In any event, you are not being asked to vary these constants, merely determine if f(x) is differentiable. You can't simply pick one part of the piecewise function and decide to use that part alone to calculate the derivative or determine differentiability.

It might be more clear if f(x) is specified as:
Code:
          x + (π/2),           for x < -π/2;
f(x) =    sin x ,              for (-π/2) ≤ x < 0;
          (tan x) / (1 + ³√x), for x ≥ 0
 
  • #3
Hello SteamKing,

Thank you for your reply! I translated it from Dutch... sorry for my mistake.
Indeed, I am trying to check if a function f(x) has a derivative at the gives values of x.

For -π/2, I choose f(x) = (sin (x)) and plug-in -π/2. It becomes >> f'(-π/2) = cos (-π/2) = 0,99 = 1

For 0, I choose (tan x) / (1 + ³√x). I suppose I should derivate using the quotient rule and then plug-in 0. But then I get 0 as result.
 
  • #4
SixBooks said:
For -π/2, I choose f(x) = (sin (x)) and plug-in -π/2. It becomes >> f'(-π/2) = cos (-π/2) = 0,99 = 1
Where does the 0,99 come from? It is wrong.
For this point you have to consider function values a both sides of -π/2 - the function is identical to the sine function at one side only.

For 0, I choose (tan x) / (1 + ³√x). I suppose I should derivate using the quotient rule and then plug-in 0. But then I get 0 as result.
You cannot "choose" anything here. And how do you get zero?
 
  • #5
Could someone solve it for me? At least I could see it and try to understand how it has to be done.
I really have no clue on how to solve this...
 
  • #6
SixBooks said:
Could someone solve it for me? At least I could see it and try to understand how it has to be done.
I really have no clue on how to solve this...
No, we won't do your work for you. From the forum rules (see Homework Guidelines at https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/)
Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.
Did you read what SteamKing and mfb wrote? They have provided some guidance for you.
 
Last edited:
  • #7
Perhaps you would find it helpful to first plot the graph. This might make the question of continuity immediately obvious. Next, you have to evaluate the derivatives of both the left side and right side functions at the point in question to see if they are the same.
If you have trouble with the derivatives, you could always use an online tool like wolframalpha.com to check your answers.
 

1. Is the function continuous at -π/2?

Yes, in order for a function to be differentiable at a point, it must also be continuous at that point. Therefore, if a function is differentiable at -π/2, it is also continuous at that point.

2. How do you determine if a function is differentiable at a specific point?

A function is differentiable at a point if the derivative exists at that point. This means that the function must be continuous at that point and the limit of the difference quotient must exist as x approaches the given point.

3. What is the difference between differentiability and continuity?

Continuity refers to the smoothness of a function and whether or not there are any breaks or gaps in the graph. Differentiability, on the other hand, is a more strict requirement and refers to whether or not the function has a well-defined slope at a specific point.

4. Can a function be differentiable at a point but not continuous?

No, a function cannot be differentiable at a point if it is not continuous at that point. In order for a function to have a well-defined derivative, it must also be continuous at that point.

5. How do you test for differentiability at 0?

To test for differentiability at 0, you can use the definition of differentiability and take the limit of the difference quotient as x approaches 0. If the limit exists, then the function is differentiable at 0. You can also check if the function is continuous at 0, as this is a necessary condition for differentiability.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
753
  • Calculus and Beyond Homework Help
Replies
7
Views
254
  • Calculus and Beyond Homework Help
Replies
26
Views
886
  • Calculus and Beyond Homework Help
Replies
1
Views
258
  • Calculus and Beyond Homework Help
Replies
21
Views
2K
  • Calculus and Beyond Homework Help
Replies
0
Views
139
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
23
Views
932
  • Calculus and Beyond Homework Help
Replies
2
Views
252
Back
Top