Inverse of a function is not differentiable

In summary: There are many other functions that would also work, but this one is simple and straightforward. In summary, the conversation discussed finding a 1-1 function that maps [-1,1] onto [-1,1] and has a derivative of 0 in the middle. Several examples were mentioned, including trigonometric and polynomial functions, with the final consensus being that f(x) = x^3 is a simple and easy-to-understand example that satisfies all the conditions.
  • #1
victoranderson
34
0
Please see attached.

I am not sure whether my example of this function is correct.

f(x) = ##sin(\frac{\pi x}{2})##
obviously, f(x) is continuous on [-1,1] and differentiable on (-1,1)

Inverse of f(x) will be ##\frac{2 sin^{-1}x}{\pi} ##

and d/dx (inverse of f(x)) will be ##\frac{2}{π \sqrt{1-x^2}}##

So not differentiable in -1,1 ??
 

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  • #2
That inverse function looks differentiable to me.
 
  • #3
anyone can help and explain?
 
  • #4
victoranderson said:
Please see attached.

I am not sure whether my example of this function is correct.

f(x) = ##sin(\frac{\pi x}{2})##
obviously, f(x) is continuous on [-1,1] and differentiable on (-1,1)

Inverse of f(x) will be ##\frac{2 sin^{-1}x}{\pi} ##

and d/dx (inverse of f(x)) will be ##\frac{2}{π \sqrt{1-x^2}}##

So not differentiable in -1,1 ??
Your problem as an image:
attachment.php?attachmentid=67141&d=1393706030.jpg


The inverse of the function you propose is differentiable on (-1,1). It's derivative does approach +∞ as x → -1 or +1 . ( Why does it approach ∞ at 1 & -1 ? -- What is it about f(x) that causes this?)

Can you think of a 1 to 1 function which maps [-1, 1] onto [-1, 1] but has a slope of zero somewhere on the interval (-1, 1) ?
 
  • #5
SammyS said:
Your problem as an image:
attachment.php?attachmentid=67141&d=1393706030.jpg


The inverse of the function you propose is differentiable on (-1,1). It's derivative does approach +∞ as x → -1 or +1 . ( Why does it approach ∞ at 1 & -1 ? -- What is it about f(x) that causes this?)

Can you think of a 1 to 1 function which maps [-1, 1] onto [-1, 1] but has a slope of zero somewhere on the interval (-1, 1) ?

Thanks for your help

so I should think of a question which has two parts?

for example f(x)=##cos(\frac{\pi x}{2})## for x=/=0. and f(x)=1 for x=0
so the inverse of this kind of function is not differentiable, right?
 
  • #6
You're overthinking this. The derivative of the inverse is 1/ the derivative of the function, so all you need is a function whose derivative is 0. In the middle, at x = 0, for example.
 
  • #7
PeroK said:
You're overthinking this. The derivative of the inverse is 1/ the derivative of the function, so all you need is a function whose derivative is 0. In the middle, at x = 0, for example.
Yes, I have been thinking this part for 3 hours...

Is the function f(x)=cosx satisfy the requirements?

f'(x)=-sinx so at x=0, f'(x)=0

derivative of the inverse = 1/sin(arcosx), so x=1 or -1, the denominator is 0

correct?
 
  • #8
cos(x) is not 1-1 on [-1, 1]

You need a function that is 1-1, has a derivative of 0 in the middle, which implies a point of inflection.
 
  • #9
victoranderson said:
Yes, I have been thinking this part for 3 hours...

Is the function f(x)=cosx satisfy the requirements?

f'(x)=-sinx so at x=0, f'(x)=0

derivative of the inverse = 1/sin(arcosx), so x=1 or -1, the denominator is 0

correct?
Is there any reason for you to be using trig functions?
 
  • #10
PeroK said:
cos(x) is not 1-1 on [-1, 1]

You need a function that is 1-1, has a derivative of 0 in the middle, which implies a point of inflection.
i m almost crazy..may be i m an idiot
i cannot think of a equation which satisfy these conditions

1. 1 to 1
2. f:[-1,1] to [-1,1]
3. derivative is 0 in the middle
4. has an inverse

i now think of e^x - x
but this kind of equation seems i m overthinking
 
Last edited:
  • #11
SammyS said:
Is there any reason for you to be using trig functions?

In my opinion trig function is the easiest way to satisfy [-1,1] to [-1,1]
 
  • #12
victoranderson said:
In my opinion trig function is the easiest way to satisfy [-1,1] to [-1,1]

You can also do it with many other functions. How about a simple power?
 
  • #13
Dick said:
You can also do it with many other functions. How about a simple power?

finally, x^3 satisfy all conditions, right?

f(x)=x^3
f'(x)=3x^2
inverse of f(x) = x^(1/3)
derivative of the inverse = 1/3 * x^(-2/3) , so not differentiable at x=0
 
  • #14
victoranderson said:
finally, x^3 satisfy all conditions, right?

f(x)=x^3
f'(x)=3x^2
inverse of f(x) = x^(1/3)
derivative of the inverse = 1/3 * x^(-2/3) , so not differentiable at x=0

Sure, that's an easy example.
 
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FAQ: Inverse of a function is not differentiable

1. What is the inverse of a function?

The inverse of a function is another function that "undoes" the original function. It is obtained by switching the input and output variables of the original function.

2. Why is the inverse of a function not always differentiable?

The inverse of a function is not always differentiable because the original function may not be one-to-one, meaning that multiple inputs can result in the same output. This results in a non-unique inverse function, which is not differentiable.

3. How can I determine if the inverse of a function is differentiable?

To determine if the inverse of a function is differentiable, you can use the horizontal line test. If a horizontal line intersects the original function at more than one point, the inverse function will not be one-to-one and therefore not differentiable.

4. Can the inverse of a differentiable function be not differentiable?

Yes, it is possible for the inverse of a differentiable function to be not differentiable. This occurs when the original function is not one-to-one and therefore does not have a unique inverse.

5. What are some examples of functions whose inverses are not differentiable?

Some examples of functions whose inverses are not differentiable include trigonometric functions such as cosine and tangent, as well as polynomial functions with an even degree. These functions are not one-to-one and therefore do not have unique inverses.

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