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Help with problem - levi98 - 08-04-2019 06:50 PM
![[Image: 7526eccb3b6995ba5314e1b39f69c806.png]](https://i.gyazo.com/7526eccb3b6995ba5314e1b39f69c806.png)
Using hp prime
need help
RE: Help with problem - Joe Horn - 08-05-2019 04:24 AM
In CAS: domain((x-5)/(x^2-25)) --> x≠-5 AND x≠5
RE: Help with problem - Aries - 08-05-2019 12:10 PM
(08-04-2019 06:50 PM)levi98 Wrote:
Using hp prime
need help
You have to use the domain() func of the calc (in CAS view)
Best,
Aries
RE: Help with problem - Marcel - 08-05-2019 12:32 PM
Hi!
For myself I prefert to do a simplification before using de domain function!
After that, the domain is OK…
Choice A with -5.
Marcel
RE: Help with problem - ijabbott - 08-05-2019 07:19 PM
(08-05-2019 12:32 PM)Marcel Wrote: Hi!
For myself I prefert to do a simplification before using de domain function!
After that, the domain is OK…
Choice A with -5.
Marcel
Technically, doesn't simplifying \(\frac{x-5}{x^2-25}\) to \(\frac{1}{x+5}\) change the domain? I know the original approaches \(\frac{1}{10}\) in the limit as \(x \to 5\), but it should probably have an open circle at \((5, \frac{1}{10})\).
RE: Help with problem - Han - 08-06-2019 06:26 PM
(08-05-2019 12:32 PM)Marcel Wrote: Hi!
For myself I prefert to do a simplification before using de domain function!
After that, the domain is OK…
Choice A with -5.
Marcel
Unfortunately, "simplifying" here is actually division, whereby \( x-5 \) in both the numerator and denominator would divide each other out EXCEPT when \( x=5 \) because this would lead to division by 0.
RE: Help with problem - Marcel - 08-07-2019 12:34 PM
Hi!
I Don't know what to say…
[attachment=7589]
Marcel.
RE: Help with problem - DrD - 08-07-2019 01:17 PM
Using the Advanced Graphing App:
[attachment=7591] [attachment=7592]
RE: Help with problem - ijabbott - 08-07-2019 01:36 PM
(08-07-2019 12:34 PM)Marcel Wrote: Hi!
I Don't know what to say…
Marcel.
Maybe this helps? 2.7 Holes in Rational Functions.
RE: Help with problem - Marcel - 08-07-2019 05:52 PM
Hi ijabbott!
Thank you, this is clear now.
Marcel.
RE: Help with problem - jlind - 08-10-2019 10:01 PM
(08-04-2019 06:50 PM)levi98 Wrote:
Using hp prime
need help
I took one look at this and realized the answer is A, and I'm with the several others who stated that X cannot be +5 or -5. Either value makes the denominator zero. When graphed by hand, it's clear that y becomes great without bound when:
x --> -5
y is therefore undefined if x = -5, as it's a real number (-10) divided by zero. The graph shows y becomes great in the negative direction as x --> -5 from less than -5 (e.g. -5.11), and it gets great in the positive direction as x --> -5 from larger than -5 (e.g. -4.99). The principle being demonstrated here is why, by definition, any number (including 0) that's divided by 0 is undefined. The direction in which it's great without bound, + or -, is indeterminate. It cannot be both simultaneously.
There must also be a hole shown where x = 5, as it creates a denominator with the original quadratic that goes to zero. By definition, 0 divided by any number other than 0 is zero. Zero divided by zero, just like any other real number divided by zero, is also undefined. In other words, 0/0 is not equal to one. I believe that's the underlying principle here. At the limit where x --> 5, y --> 0.1, but that's not the same as x = 5. Plug in x = 5.1 and 5.01, and then x = 4.9 and 4.99 to see what you get. It converges on each side of 0.1 and there's an infinitesimally tiny hole for exactly x = 5. Zero is a special case that is neither a positive nor a negative real number. It has no "sign" like all other real numbers. Hope this explains some of the underlying principles surrounding why division of any real number, including zero, by zero is undefined.
John
RE: Help with problem - Aries - 08-12-2019 05:58 AM
There is one vertical asymptote (x=-5) and y=0 (i.e. x-axis) is the horizontal asymptote.
There is no slant asymptote.
Best,
Aries
RE: Help with problem - Tonig00 - 08-13-2019 04:36 PM
I would likevto point out that is clearvthat 0/0 is undefined.
But any real number>0 divided by 0 I would say is infinite.
When we have a function in single variable I would say that if left an right limits are the same, then it is definite.
It may be the case of for example
5/(x-5)^2.
RE: Help with problem - jlind - 08-14-2019 05:38 AM
(08-13-2019 04:36 PM)Tonig00 Wrote: I would likevto point out that is clearvthat 0/0 is undefined.
But any real number>0 divided by 0 I would say is infinite.
When we have a function in single variable I would say that if left an right limits are the same, then it is definite.
It may be the case of for example
5/(x-5)^2.
Tonig00,
Any number, divided by zero, is undefined.
- "Infinity" isn't a number, it's a concept. There's no such thing as Infinity+1, or Infinity*3 or Inifinity^2 being greater than Infinity.
- For any number, a <> 0, there is no number b for which 0 * b = a, which must exist by the definition of multiplication and division. 0 * b = 0, for any number b, but a <> 0. This is an impossibility.
Let 1 = x
Multiply by x to get
x = x^2
Subtract 1 from each side to get
x - 1 = x^2 - 1
Divide both sides by x − 1 (this is a hidden division by zero as x = 1)
(x - 1) / (x - 1) = (x^2 - 1) / (x - 1)
1 = ((x - 1) * (x + 1)) / (x-1)
which simplifies to
1 = x + 1
Since x = 1, by substitution:
1 = 1 + 1, and therefore:
1 = 2
This is impossible. Hope this helps some with understanding why any number divided by zero is undefined.
John
RE: Help with problem - Aries - 08-14-2019 06:18 AM
(08-13-2019 04:36 PM)Tonig00 Wrote: I would likevto point out that is clearvthat 0/0 is undefined.
But any real number>0 divided by 0 I would say is infinite.
When we have a function in single variable I would say that if left an right limits are the same, then it is definite.
It may be the case of for example
5/(x-5)^2.
I'd say 0/0 is indeterminate (form), taking limit into account.
Best,
Aries