this is a question assuming that "g" is a given function
2006-09-26 08:29:05 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Engineering
Are you talking about an inverse function? There isn't necessarily anything wrong with that. However, it should be noted that not every function has an inverse function.
If a function returns the same y-value for two different x-values, then it doesn't have an inverse function. The inverse would have to return two different y-values for the same x-value, and functions aren't allowed to do that.
2006-09-26 08:33:40 · answer #1 · answered by Bramblyspam 7 · 0⤊ 0⤋
Depends on how it is written. g^-1(x) would be the inverse function of x. E.g. g(x)=sin(x), g^-1(x)=sin^-1(x) If it is written g(x)^-1, it simply means the reciprical, 1/g(x). Division by zero is an issue here, but not always a problem. You just have to realize that the limit oft the function does not exist (goes to infinity) at the point g(x)=0. This won't cause any problems if the situation does not involve this point. There are also other operations which cannot be performed on this equation without restrictions.
2006-09-26 17:35:57 · answer #2 · answered by awakeatdawn 3 · 0⤊ 0⤋
g to the power minus 1 means 1/g.
Division by zero is not allowed (the division by zero police will be at your doorstep - don't do it!)
Since g is a function that can take on different values,
1/g has a problem if g=0.
2006-09-26 14:01:10 · answer #3 · answered by WildOtter 5 · 0⤊ 0⤋
it usually indicates an inverse function.
2006-09-26 08:36:43 · answer #4 · answered by Weasel 4 · 0⤊ 0⤋
none, the minas one just means that if g=2x
g^-1=1/2X
2006-09-26 10:00:43 · answer #5 · answered by darren p 2 · 0⤊ 0⤋
always with negative waves....think positive
2006-09-26 08:47:39 · answer #6 · answered by Anonymous · 0⤊ 0⤋