Inverse of a function is not differentiable

[victoranderson]

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 ??

 

[PeroK]

That inverse function looks differentiable to me.

 

[victoranderson]

anyone can help and explain?

 

[SammyS]

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:

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) ?

 

[victoranderson]

Your problem as an image:

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?

 

[PeroK]

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.

 

[victoranderson]

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?

 

[PeroK]

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.

 

[SammyS]

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?

 

[victoranderson]

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

 

[victoranderson]

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]

 

[Dick]

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?

 

[victoranderson]

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

 

[Dick]

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.