Continous and differentiable function

[jaus tail]

Homework Statement

Homework Equations

All polynomial functions are continuous so the function is continuous everywhere.
For differnetiable we differentiate the polynomial. But how to do this?

The Attempt at a Solution

I only removed option A as polynomial functions are always continuous. How to check for differentiable here? In solved examples the function is 1 function and not max or min.

 

[fresh_42]

Polynomials are also differentiable. The point is, this isn't a polynomial function, it is composed of polynomial functions. So you have to consider the points at which one function is clued to the other function: Is there a step (gap)? Is it continuous at those points? Is it differentiable there, and why or why not?

 

[Delta2]

Consider the equation $1-x=x^2-1$ , the roots are the points where the function changes from one polynomial to the other. In between the roots which polynomial is bigger? So after you answer that, just check for continuity and differentiability at those points by computing the left and right limits and the left and right derivatives.

 

[epenguin]

Sketch it! Sketch f = (x - 1) and f = (x² - 1) and on top make the actual given function bold. You might have some trouble formalising a proof argument, but at least you won’t have any doubt what the answer is.

 

[jaus tail]

If I take 1-x = x² - 1 Roots of this equation are -2 and 1
For x = -2
1 - x = 1 + 2 = 3
x² - 1 = 4 - 1 = 3

For x = 1
1-x = 0
x² - 1 = 0

Between roots. So I take point x = -1
Now:
1 - x = 1 + 1 = 2
x² - 1 = 1 - 1 = 0
So 1 - x is bigger.

Plotting graph i get this:

Now what to do? I don't even understand the options.

 

[jaus tail]

It looks continuous everywhere. and even differnetiating is easy. 1 - x derivative is -1, and x^2 - 1 derivative is 2x.
I get option B but answer option is C.

 

[epenguin]

By ‘on top’ I meant overlaying. Just the first two curves, and then the parts of them that corresponds to your function you make more bold, thicker.

 

[fresh_42]

Let $f(x)=1-x\, , \,g(x)=x^2-1\; , \;m(x)=max\{f(x),g(x)\}$. It's better to have different names, not just $f$.

It looks continuous everywhere...

Yes. No gaps or anything. Formally you have to show that $f(-2)=g(-2)$ and $f(1)=g(1)$. The other areas are polynomials, so they are both, continuous and differentiable there, as you mentioned:

... and even differnetiating is easy. 1 - x derivative is -1, and x^2 - 1 derivative is 2x.

Before you draw conclusions, what happens at $x=-2$ and at $x=1\,?$ Gaps? Vertices? And what does this mean?

I get option B but answer option is C.

 

[Delta2]

It is continuous everywhere. But at x=-2 the right derivative is -1 but the left derivative is 2x=-4. At x=1 the left derivative is -1 but the right derivative is 2x=2.

 

[jaus tail]

It is continuous everywhere. But at x=-2 the right derivative is -1 but the left derivative is 2x=-4. At x=1 the left derivative is -1 but the right derivative is 2x=2.

Yeah makes sense.

 

[jaus tail]

Let $f(x)=1-x\, , \,g(x)=x^2-1\; , \;m(x)=max\{f(x),g(x)\}$. It's better to have different names, not just $f$.
Yes. No gaps or anything. Formally you have to show that $f(-2)=g(-2)$ and $f(1)=g(1)$. The other areas are polynomials, so they are both, continuous and differentiable there, as you mentioned:
Before you draw conclusions, what happens at $x=-2$ and at $x=1\,?$ Gaps? Vertices? And what does this mean?

To be honest I said it's continuous everywhere cause it looks so in graph. I didn't really understand what it means. I mean i know that if a function has sharp turns then it's discontiuous. But what do gaps mean here?

Of the three functions above, only blue one is continuous right?
What about yellow one? It's y = mod x

 

[member 587159]

No you are confusing differentiability and continuity. If the graph looks connected (i.e. we can draw it without lifting up our pencil), it is continuous. If in a point, there is a sharp turn, it is not differentiable in that point.

 

[fresh_42]

The thumb rule is: if you can draw it without lifting the pencil, then it's continuous. (Of course there are pathological functions where this is not true, but here it is.) I recommend to look up the formal definition of what continuity means, e.g. on Wikipedia. Then you should try to formally prove continuity of $m(x)$ at e.g. $x=-2$.

The thumb rule for differentiability is to call it smooth. (This is not really correct, as smooth normally means infinitely often differentiable, and differentiable means only once.) Vertices and sharp turns are NOT smooth, because direction changes from one moment to the other with no in-between. Again, you should look up differentiability and try to prove, that $m(x)$ is NOT differentiable at, e.g. $x=-2$. @Delta² basically already gave you the answer, but I think you should look it up, e.g. on Wikipedia.

 

[jaus tail]

Thanks. till now I thought step function is not continuous as at t = 0 there is abrupt rise. thanks for clarifying it. will look up in google for examples.

 

[Dick]

Thanks. till now I thought step function is not continuous as at t = 0 there is abrupt rise. thanks for clarifying it. will look up in google for examples.

You are right that the step function is NOT continuous. You have to be careful about 'drawing without lifting the pencil'. The graph of the step function does not have vertical lines in it. Functions in general don't. Do you see why?

 

[jaus tail]

You are right that the step function is NOT continuous. You have to be careful about 'drawing without lifting the pencil'. The graph of the step function does not have vertical lines in it. Functions in general don't. Do you see why?

Cause functions have slope. step function has zero slope but at t = 0, it has infinite slope.

 

[Dick]

Cause functions have slope. step function has zero slope but at t = 0, it has infinite slope.

Sort of. But it's not so much about slope. Functions can only have one y-value corresponding to any value of x. So vertical lines are out.

 

[jaus tail]

Okay so this clears up continuous that the line must be smooth and have unique y values. And for differentiability the left and right side values must be same, right?
But then a sine wave, at 90 degrees has different slopes on both sides. It has positive slope on left side and negative slope on right side?
So shouldn't sin (theta) be non differentiable at theta = 90 degrees?

 

[fresh_42]

But then a sine wave, at 90 degrees has different slopes on both sides.

Why that? It has different slopes on each side, positive and negative and zero in the middle. So it is a continuous move from decreasing positive values, to zero, to increasing negative values. The limits from the left and from the right are the same, namely $0$. Both continuity and differentiability are limit processes.

It has positive slope on left side and negative slope on right side?
So shouldn't sin (theta) be non differentiable at theta = 90 degrees?

Edit: To be more exact: they are local properties, i.e. they have to hold in small neighborhoods of a point.

 

[jaus tail]

But then in post#9 the derivatives also are different. Even for sine wave the derivatives are different. ??

 

[fresh_42]

But then in post#9 the derivatives also are different. Even for sine wave the derivatives are different. ??

There is a difference: We have $\lim_{x \to -2 -0} m'(x) = \lim_{x \to -2 -0} g'(x)= -4 \neq -1 = \lim_{x \to -2 +0} f'(x) = \lim_{x \to -2 + 0} m'(x)$ whereas $\lim_{x \to 90° -0 } (\sin)' x = \lim_{x \to 90° -0 } \cos x= 0 = 0=\lim_{x \to 90° +0 } \cos x= \lim_{x \to 90° +0 } (\sin)' x$ so there is no difference in the approach from the left or from the right on the sine function. However, it makes a difference whether you come from the left or from the right on the given maximum function $m(x)$.

 

[epenguin]

The orange and blue parts of your #5 you just needed to bold the parts of them that correspond to the definition of f(x) in the first line of your question. Plus maybe some labels on coordinate axes for some significant points.

Congratulations on using some app or or or other that gives clear pictures instead of the usual dark yellow-grey upsidedown homework submissions!

 

[jaus tail]

Okay got it. the derivative must be unique and must exist for function to be differentiable. And for continuous we just limits to be same on left side and right side.

 

[jaus tail]

I didn't use app. I used paint software. Actually i have an exam in 10 feb on electrical engineering. I'm giving practice tests for that.