Re: Composite function help (sine and signum)

From: Ivan (ivan999_at_remove34this.iprimus.com.au)
Date: 03/06/05 

Date: 7 Mar 2005 00:58:59 +1100

A N Niel <anniel@nym.alias.net.invalid> wrote in
news:060320050717145202%anniel@nym.alias.net.invalid:

> signum is defined in your book? If it were not, I would have said
> signum(x) = 1 for x>0, = -1 for x<0 and (perhaps) = 0 for x=0 .
>
> In article <Xns9611A8C6E3726ivan999remove34thisi@203.134.67.67>, Ivan
> <ivan999@remove34this.iprimus.com.au> wrote:

<SNIP>

>> Given:
>> f(x)=sin(x), g(x)=SIGNUM(x), h(x)=sqrt(x), k(x)=intg(x)
>>

<SNIP>

>> (d) Determine g{f[h(x)]} and state the domain and range
>> ....... More trouble!
>>
>> sqrt(x) is defined for all x>=0. If sin(x) is defined for all x>=0
>> and has the range {y real, 0<=y<=1} then the domain of the composite
>> function is {x real, 0<=x<=1} and the range is {y integer, 0<=y<=1}.
>>
>> Is this correct?
>
> Not with my definition of signum, above.
>

Thanks for the reply and we agree on the definition of signum!

Let me try again ...

g{f[h(x)]} = SIGNUM{sin[sqrt(x)]}

sqrt(x) has the domain {x real, x>=0} and range (y real, y>=0}
Therefore, taking the range of the previous function to be the domain of
the next, sin[sqrt(x)] has the domain {x real, x>=0} and range {y real,
-1<=x<=1}
Finally then, SIGNUM{sin[sqrt(x)]} has the domain {x real, -1<=x<=1) and so
has the range {y integer, -1<=y<=1}.

Since you didn't comment on the other stuff I assume it was correct :)

Thanks for your help - hopefully we agree on this last point now?

Ivan.

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