A continuous function may not be differentiable

[Maddy_Math]

How to prove "A continuous function may not be differentiable"
graphical case is; when there is vertical tangent but how to do in {non-graphical or algebraic calculus or I don't know what exact term to use}

 

[pka]

How to prove "A continuous function may not be differentiable"
graphical case is; when there is vertical tangent but how to do in {non-graphical or algebraic calculus or I don't know what exact term to use}

All you have to do is give an example: $\displaystyle f(x)=|x|$.
That function is continuous at $\displaystyle x=0$ but not differentiable.

 

[Maddy_Math]

All you have to do is give an example: $\displaystyle f(x)=|x|$.
That function is continuous at $\displaystyle x=0$ but not differentiable.

but how to prove in calculus without graph that derivative of |x| at x = 0 does not exist

 

[stapel]

but how to prove in calculus without graph that derivative of |x| at x = 0 does not exist

One method might be to convert the absolute-value form of the function into the piecewise form (with one rule for x < 0 and another for x > 0). Differentiate the halves of this form.

What do you see? What do these results tell you about the derivative at x = 0? :wink:

 

[lookagain]

but how to prove in calculus without graph that derivative of |x| at x = 0 does not exist

Suppose

$\displaystyle f(x) \ = \ |x|$

$\displaystyle f(x) \ = \ \sqrt{x^2}$

$\displaystyle f(x) \ = \ (x^2)^{\frac{1}{2}}$


$\displaystyle f'(x) \ = \ \frac{1}{2}[(x^2)^{-\frac{1}{2}}](2x)$


$\displaystyle f'(x) \ = \ \dfrac{x}{(x^2)^{\frac{1}{2}}}$


$\displaystyle f'(x) \ = \ \dfrac{x}{\sqrt{x^2}}$


$\displaystyle f'(x) \ = \ \dfrac{x}{|x|} \ \ \ \ \bigg(or \ \ \ \dfrac{|x|}{x}\bigg)$

 

[mmm4444bot]

A continuous function may not be differentiable

graphical case is; when there is vertical tangent

Vertical tangent is one case; there are other cases, too.

 

[Maddy_Math]

Suppose...........

Yeah that's it thankyou "lookagain" I got it

 

[Maddy_Math]

Vertical tangent is one case; there are other cases, too.

cases other than vertical tangent are not continuous I guess or am I wrong (I'm not sure)

 

[HallsofIvy]

cases other than vertical tangent are not continuous I guess or am I wrong (I'm not sure)

You are wrong and the examples already given show that: f(x)= |x| is continuous for all x but is not differentiable at x= 0. There is no vertical tangent at x= 0- there is no tangent at all.

 

[Maddy_Math]

You are wrong and the examples already given show that: f(x)= |x| is continuous for all x but is not differentiable at x= 0. There is no vertical tangent at x= 0- there is no tangent at all.

Oh thanks HallsofIvy, well I hope there aren't any other cases than these two

 

[mmm4444bot]

I hope there aren't any other cases than these two

Why? Too much to remember? :wink:

There is a third case.

A continuous function that oscillates infinitely at some point is not differentiable there.

Look at the graph of f(x) = sin(1/x). This function is continuous at x=0 but not differentiable there because the behavior is oscillating too wildly.

If you would like a reference sheet of function types (both continuous and with discontinuity) that have places which are not differentiable, you could print out this page.

Cheers :cool:

 

[Maddy_Math]

thanks m4bot I got it

 

[HallsofIvy]

Why? Too much to remember? :wink:

There is a third case.

A continuous function that oscillates infinitely at some point is not differentiable there.

Look at the graph of f(x) = sin(1/x). This function is continuous at x=0 but not differentiable there because the behavior is oscillating too wildly.

No, that function is not continuous at x= 0- it is not even defined there. And the limit, as x goes to 0, does not exist so you cannot define it at 0 to make it continuous. Perhaps you meant f(x)= x sin(1/x) if x is not 0, f(0)= 0.

If you would like a reference sheet of function types (both continuous and with discontinuity) that have places which are not differentiable, you could print out this page.

Cheers :cool:

 

[Quaid]

No, that function is not continuous at x= 0-

Whoops. You are correct, of course; we would need to remove the discontinuity at zero, by piecewise definition, to make a continuous function.

I mistakenly posted that function thinking (without thinking) that it was already continuous. :roll:


Here's a different situation: the Weierstrass function. Continuous everywhere; differentiable nowhere.