TT's Maths Thread

[/0]

hmm maybe

[kamil9876]

if we can prove that for an arbitrarily small then we are done.

It is based on the
Hint: let

edit: actually i think it only works if

Though i think letting is a good idea, similair idea to /0's.

[kamil9876]

yeah here is a completion of /0's start:

provided

So we need: and to be satisfied simultaenously.

Hence we need: (1)  and (2) to be satisfied simultaenously.

(1) and (2) are satisfied simultaenously if:





which is satisfied if

so we found the


edit: some small mistake. Hope there is a neater and general way for all continous and invertible functions.

[TrueTears]

haha awesome, yeah the crux was finding the minimum, the rest is trivial xD

[TrueTears]

How do I show is continuous in its domain and find it's domain.

So I split it up into 3 functions.







So

I know that if the functions that make up the composite function is continuous then so is the composite, but how do I apply that here and prove it and find the domain that it's continuous over?

[kamil9876]

should be the domain.

This is the maximal domain for which is continous(exclude 0 since in that case the limit doesn't exist, (no limit from negative side)). It has range

therefore is continous since is continous if the domain of is . But the range of of this function is But is continous if the domain of is .

[TrueTears]

Thanks.

What about this one:

Locate the discontinuities in the following function:



Let



First we need ensure is defined.





So

Now we need to ensure is defined.

So

But how to find ?

[/0]

where

For all defined x,   , so it's decreasing in its continuous sections.

Since there is an asymptote at x = 0,

We can check and for the maximums
and and for the minimums.

[TrueTears]

where

For all defined x,   , so it's decreasing in its continuous sections.

Since there is an asymptote at x = 0,

We can check and for the maximums
and and for the minimums.

Ah thanks, could you also work it out this way? (It's the way Stewarts suggests)

can take a range of values from

Now let

We have

So when you sub in we get a very small value but it will approach 0.

We can't have

When you sub in we have is very large, so it will also approach .

Thus the range of is

[TrueTears]

and are positive numbers, prove that the equation has at least root in the interval .

Since both are continuous then their sums are also continuous.

Let

Thus using the intermediate value theorem if we can find a such that then there must be at least 1 root.

So if then there must be a .

But how can I evaluate ? is undefined at .

[/0]

You don't need to input -1 and 1, you can try -0.9 and 0.9

[TrueTears]

You don't need to input -1 and 1, you can try -0.9 and 0.9

lol, but those numbers are so ugly

[TrueTears]

Show using the squeeze theorem that

How would you do this using the squeeze theorem? =S

I can do it without using the theorem...

[/0]

hmm this might work

for all

[TrueTears]

hmm this might work

for all

pr0pr0pr0pr0pr0!!!!!