can anyone elxplain this calculus problem to me please?

[crawdaddy029]

http://tinypic.com/view.php?pic=ape5mu&s=6

thanks.

[Arkhias]

Read through this:

http://oregonstate.edu/instruct/mth2...iffVsCont.html

Make sure to click the "Why is that?" links.

[braeldiil]

In simple terms, it's not differentialable at x=-5 or x=-1 because it has discontinuities at those points. Even though it is continuous, it's also not differentiable at -3. This is because it has a "point", or more generally, because the curve is not smooth. If you come in from the left, the curve has a slope of -2 as it approaches -3. If you come in from the right, the curve has a slope of 1 as it approaches -3.

[Spectral]

Put in non-mathy terms (I guess this is dopey phrasing, but whatever), the points that a function won't be differentiable are the points where the function doesn't have a determinant rate of change. In this case, that's the two discontinuous points, and the point where the line changes direction). I hope that explanation makes a tiny bit of sense... my math is so rusty.

[crawdaddy029]

In simple terms, it's not differentialable at x=-5 or x=-1 because it has discontinuities at those points. Even though it is continuous, it's also not differentiable at -3. This is because it has a "point", or more generally, because the curve is not smooth. If you come in from the left, the curve has a slope of -2 as it approaches -3. If you come in from the right, the curve has a slope of 1 as it approaches -3.

ok thanks, I don't know dick about calc so that first post didn't help me any I am electrician so I know formulas but calc is not a strong point. anyway can I get help with this one now please.

http://tinypic.com/r/2yovde8/6

[Trubo]

Been quite some time since I've taken Calculus. If that second problem is asking you to find what F PRIME of X is at X = 0, I want to say it's 0 as F(x) when x = 0 is simply 0, and the derivative of 0 is still 0.

[simplemind]

Been quite some time since I've taken Calculus. If that second problem is asking you to find what F PRIME of X is at X = 0, I want to say it's 0 as F(x) when x = 0 is simply 0, and the derivative of 0 is still 0.

can you show work? I kind of understand what you're saying but I have to show work.

[Deleted]

Been quite some time since I've taken Calculus. If that second problem is asking you to find what F PRIME of X is at X = 0, I want to say it's 0 as F(x) when x = 0 is simply 0, and the derivative of 0 is still 0.

My calculus is terriby rusty, but i think that f(x) is not differentiable at 0 in the second problem.

If i'm doing things right, outside 0, f'(x) = 2 x sin(5/x) - 5 cos(5/x)
when we approach 0, the first term goes to zero, so f'(x) when x->0 = -5cos(5/x), which is indeterminate at x=0*.
Hence, f'(x) is non-continuous for x=0, which means f(x) is not differentiable for x=0.


* cos(1/x) is a sinusoidal function of frequency tending to infinite as it approaches zero: you can see quite easily why it's indeterminate.

PS.- Please, verify that i did the derivative correctly XD

[Coombs]

can you show work? I kind of understand what you're saying but I have to show work.

There isn't really much work to show unless they want you to solve for both derivatives and then use limits of infinity to rule out the sin and cos that come from the derivative. It has been a long time since I had to deal with this but I assume it's as easy as saying that the derivative does exist at f'(0). I'm not sure if the function stops working when the x variable is no longer present, I can say though that the first function x cannot be zero.

---------- Post added 2012-09-28 at 03:17 AM ----------

My calculus is terriby rusty, but i think that f(x) is not differentiable at 0 in the second problem.

If i'm doing things right, outside 0, f'(x) = 2 x sin(5/x) - 5 cos(5/x)
when we approach 0, the first term goes to zero, so f'(x) when x->0 = -5cos(5/x), which is indeterminate at x=0*.
Hence, f'(x) is non-continuous for x=0, which means f(x) is not differentiable for x=0.


* cos(1/x) is a sinusoidal function of frequency tending to infinite as it approaches zero: you can see quite easily why it's indeterminate.

PS.- Please, verify that i did the derivative correctly XD

The derivative is right and I agree with the rest. I'm just not sure if f(x)=0 when x=0 bears any significant meaning.

[Deleted]

The derivative is right and I agree with the rest. I'm just not sure if f(x)=0 when x=0 bears any significant meaning.

I think it's there to give you assurance that the f'(x) is indeed non-continuous at x=0.

I mean, if f(x) = x^2 sin(5/x) everywhere, you could argue that for x=0 the function is still continuous and derivable (because sin and cos are "nice" functions lol), even if the value of the function and the derivative for x=0 is indeterminate.
Setting in stone that f(0) = 0 removes that possible doubt.

[worcester]

for why they added f(x)=0 for x=0
(1) if we say that f(x) = 0 when x=0 is in so that the function remains piece-wise smooth (so it is defined for all x), then it insinuates that x^2 sin(5/x) is not differentiable at 0.
Which is the logic that jotabe thought. which is close, since its actually the converse logic we want:

Conversely to (1), you can see by the math that x^2 sin(5/x) is not differentiable at 0, hence you can conclude that indeed f(x) was defined at 0 so that it does remain a piece-wise smooth function (explicitly defining 0 so that the domain is R instead of the awkward R-{0}; i.e. "nice" function).]

ps. this has nothing to do with the actual maths, which is all correct :P

[Deleted]

Yes, it can go both ways. Getting a function that is defined for all R might lull the student into a false sense of security, hehe.