If
- $y = x^2$, then $y$ grows quadraticaly with $x$,
- $y = \log x$, then $y$ grows logarithmicaly with $x$.
How does $y$ depend on $x$ if $y = \sqrt{x}$?
Asked
Active
Viewed 41 times
1
Antoine
- 3,439
-
1It grows as the square root.. There is no particular name for that rate of growth. – Ethan Bolker Nov 07 '16 at 12:54
-
3I'd say that $y$ grows as the square root *of* $x$. – Bobson Dugnutt Nov 07 '16 at 12:57
-
Can one of you two post this as an answer, please? – Antoine Nov 07 '16 at 13:47
-
The $x$ grows quadratic-ally with $y$. – Simply Beautiful Art Nov 07 '16 at 14:40
-
@SimpleArt I was aware of this "solution", but if $y$ is the dependent variable, this answer does not make much sanse ... – Antoine Nov 07 '16 at 14:45
-
Sure it does, because $x$ could be the dependent variable of $y$. :P – Simply Beautiful Art Nov 07 '16 at 14:48
1 Answers
0
As the first comments by Ethan Bolker and Lovsovs suggest, the answer is that
$y$ grows as the square root of $x$.
Antoine
- 3,439