sparrow2
  • sparrow2
if x is a variable then does it have infinite number of values?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Depends on the context, for mathematics then mainly yes, for real world/physics applications there are limits.
anonymous
  • anonymous
this is the question im assigned: Does the domain change uner the following transformation of y-log x? Explain. (vert stretch by 5 and horiz. stretch by 2)
anonymous
  • anonymous
*under

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anonymous
  • anonymous
**and its y=log x . sorry
anonymous
  • anonymous
domain changes from the horizontal stretch, increases the x values that maps to y values
anonymous
  • anonymous
by stretch by 2 im guessing you mean expand and not divide by 2
anonymous
  • anonymous
yeah its expanding horizontally by a factor of 2
anonymous
  • anonymous
oh wait sorry as a log graph it won't change the domain
anonymous
  • anonymous
THANK YOU <3
Study_together
  • Study_together
It depends but I still think its values are infinite. Please click on the best response if you like the post.
anonymous
  • anonymous
what if its limited to an equation with limited solutions etc.
anonymous
  • anonymous
then it becomes limited
ganeshie8
  • ganeshie8
Define \(x\) as the variable that represents solution to the quadratic equation \[\large x^2-2x+1=0\] how many values does \(x\) have ?
anonymous
  • anonymous
Factorised into (X-1)^2=0, 1 solution
sparrow2
  • sparrow2
x=1 is solution and it is "uknown"
SolomonZelman
  • SolomonZelman
if x is a variable then does it have infinite number of values? YES, I would say. And here are some examples of why I say this. -------------------------------------------- Take any logarithmic function: \(\large\color{black}{ \displaystyle y=\log(x+a)+c }\) (with any side shift a AND any vertical shift c) Note: The domain is not impacted by C (and nor is the range since it goes infinitely up and down). Now, the log is only defined in this case, if \(\large\color{black}{ \displaystyle x+a>0 }\), and thus: \(\large\color{black}{ \displaystyle x>-a }\), and knowing that no restriction on how large (x+a) can get your domain is: (-a, +∞) So if x is over the interval (-a, +∞), then (regardless of the value of a), x takes on infinite number of values. -------------------------------------------- Obviously, any polynomial is known to be continuous over interval (-∞,+∞). This way for any polynomial function \(\large f(x)\), you get that the domain of y=f(x) is (-∞,+∞) So, here again x takes on infinite number of values. -------------------------------------------- Consider any function!! and you will either get that the domain is `(-∞, +∞)` OR `(s, +∞)` OR `(-∞, s)` (where s is some number)
SolomonZelman
  • SolomonZelman
So unless you are taking about a function that has a restriction of a≥x≥b, or some sort of limitation fron x to go infinitely to the left or right, (then) x takes on an infinite number of values.
SolomonZelman
  • SolomonZelman
Usually though, if you are taking a limit of the function, such that: \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~a}f(x)}\) \(\Huge\color{blue}(\) or, \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~a^+}f(x)}\) or, \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~a^-}f(x)}\) \(\Huge\color{blue})\) then, it would have a value, unless you're going into an undefined direction like: \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~0^+}\log_{a}(x)}\) \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~a}\frac{c}{x-a}}\) (for left-sided, right-sided or two sides limit DNE) \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty}ax^{b}}\) (for b>0) (or other examples where you meet asymptotes or just an infinite growth of x)
SolomonZelman
  • SolomonZelman
igtg
anonymous
  • anonymous
I was just answering the first main question, unless a variable is defined as being part of a function only then it would be infinite, but if it is an unknown in an equation when it can be solved for then it is limited to the value of those solutions.

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