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massive-war

Need help in Calculus (simple problem)

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massive-war    100
We're learning how to apply derivitives to active programming, however, this problem is really frustrating, and I've tried just about everything. If you're willing to help, here it is:
Is there any value of b that will make g(x) = {x + b, x < 0}  continuous at x=0?
                                              {cos(x), x>=0}
Is there a value that will make g(x) differentiable at zero?
The double stacked statements are meant to be in one big bracket, showing that they are both related to the function g(x) I just can't seem to get what the answer would even relate to, or how you would go about doing that. It's just hard because our teacher gives us a test on something we haven't learned yet, then teaches us about it. His philosophy is to make us do horrible (basically fail) so we're tempted to learn how to get better. How we're supposed to be motivated after failing something, I don't know. But all I know is that I'm motivated enough right now, as it is, to come online and ask for help because I will not fail that class, especially when I can program better than anyone else (by the way the class average is a 56) I'm not asking for sympathy or your pity, I'm just stating why I'm asking for help on a homework assignment, because I'm the same way: do it yourself bitch Thanks, Massive-war

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Zipster    2365
Best way is to draw some pictures. You have two functions, a sinusoid and a line, and they have to be continuous at x = 0. See if you can't come up with a diagram for that.

As for whether or not the function is differentiable at x = 0, that means that the derivative exists at that point. Once you find the first part, this shouldn't be too hard.

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JohnBolton    1372
The important questions are:

1. Do you know what "g(x) = {x + b, x < 0}" notation means?
(cos(x), x >= 0}
2. Do you know what "continuous" means?
3. Do you know what "differentiable" means?

You need to know what these three things mean in order to find the answer.

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Guest Anonymous Poster   
Guest Anonymous Poster

Take a look here:

http://archives.math.utk.edu/visual.calculus/1/continuous.5/
http://archives.math.utk.edu/visual.calculus/1/continuous.3/index.html

The last drill problem will help you. Remember to approach the limit from both sides.

Differentiablity will be similiar, however apply the limit definition of a deriviative.

You need to get used to questions like this... Know the formulas that are behind these definitions. They often re-appear on tests. For instance he or she might ask a question with 3 sub-domains.
http://web01.shu.edu/projects/reals/cont/derivat.html

The teacher isn't doing this to be mean, but to push you guys a bit. An effective strategy then would be to read ahead. Math isn't a spectator sport, a good teacher helps but is largely irrelevant, it's all up to you and the amount of effort you put in. Don't depend on him or her to provide you motivation.

Good Luck.





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BS-er    181
Well they're just yes/no questions, so you have a 50/50 chance with each if you just guess :). J/k, I know more is expected.

Let us know whan you have an answer. I think I know, but sometimes I only know enough to be dangerous :).

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butterscotch    122

yeah there is just one value of b that will make it continuous and that is b=1.

sorry, there is no value of b that can make g(x) differentiable at x= 0. :(


To make it continous make sure cos(x) at x= 0 i.e. cos(0) which is equal to 1 is equal to x + b at x =0, i.e. 0 + b = 1

For makin them differentiable at x = 0 , first they need to be continous there, which they luckily can be made by making b = 1,

now check if dy/dx of both the curves ( y = x + b , and y = cos(x) )at x = 0 are equal.

for the first one dy/dx = 1 , for the second dy/dx = - sin(x) ,

Now at x = 0

dy/dx of first curve (y = x + b ) = 1

dy/dx of second curve (y = cos(x) ) is -sin(0) = 0

dy/dx of the first curve actually cannot be changed using the value of b...there is no way we can make them differentiable at x = 0.


Thanks
Aditya

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grhodes_at_work    1385
Massive-war,

I sympathize with your situation, and wish you the best of luck. Have you and the other students tried talking to the teacher, or talking to another teacher that has a better rapport with students?

In any case, unfortunately technical problems from school are off topic here, and against forum policy. I'm sorry. I have to close the thread. You can read about the policy here: Forum FAQ

I don't want to leave you hanging. If you look at the FAQ, you will find links to other forums on the web that specifically focus on homework/schoolwork help. This other forums are the appropriate places for a post such as yours. I'm sure you will be able to find some help there.

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