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M K.

asked • 11/09/13

When finding the inverse of f(x)=v(x-3) (a one to one function) and stating the domain and range of the inverse, why is the domain of the inverse switched?

So if f(x)=√(x-3), I know the original domain would be [3, ∞) and the range would be [0,∞), but what would the domain and the range be for the inverse be, which is f-1=x2+3, and why would it be that way? My teacher told us the domain of the original one to one would be the range of the inverse and the range of the original would the the domain of the inverse, but wouldn't the domain be (-∞, ∞) because any integer can be squared and the result be real? And if you drew the graph, the parabola would extend to both sides of the x axis forever, right? So why would the domain be [0, ∞)?
 
I may have follow up questions about this topic too! Thanks for your help!

1 Expert Answer

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Vivian L. answered • 11/09/13

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M K.

Thanks so much! That makes sense!
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11/09/13

Vivian L.

One more thing...
The definition of a function is that each value of x is represented only once.
The rule for the representation of y is NOT a mirror image as any value can be repeated without limit.
 
Therefore, the rule as your teacher presented that the inverse of an equation would present a mirror of domain/range to range/domain, cannot be true.
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11/09/13

M K.

I'm a little confused with what you mean by "The rule for the representation of y is NOT a mirror image as any value can be repeated without limit." What does this mean?
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11/09/13

Vivian L.

Function rule for x---each value only represented once.
Function rule for y---each value can be presented without limit as to quantity.
 
These rules are not mirror images of each other.  So when x and y becomes x-1 and y-1, both x and x-1 must follow the first rule.  That is not a mirror.
 
For your teacher to say that the f(x) has a domain/range, and f-1(x) has a mirror domain/range, is impossible.
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11/09/13

M K.

When we were working these problems out in class, we had an example m(x)=(1-3x)/(2+x), x not equal to -2, and this was an example where the inverse and original function had mirror domains and ranges. So the inverse would equal (1-2x)/(x+3), the domain for the original would be (-∞, -2) U (-2, ∞), the range for the original would be (-∞, -3) U (-3, ∞), the inverse domain would be (-∞, -3) U (-3, ∞) because the denominator can equal anything but -3 without the denominator being 0 and undefined, and the inverse range would be (-∞, -2) U (-2, ∞) because the horizontal asymptote would be -2x/x which would equal -2. Then how, if what you said stands, would this be possible?
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11/09/13

Vivian L.

The class example does not involve any exponential such as x2, the exponent being an even number.
As you already know, in such circumstance, a positive and negative number will produce the same result.
For example,
y=x2
x=-2, y=4
x=2, y=4
The y-coordinate of 4 is represented twice in this function.
Each x-coordinate is only represented once in this function.
A mirror of domain-to-range, range-to-domain is impossible.
However, a mirror of absolute values of domains and ranges, is possible.
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11/09/13

Vivian L.

The example your instructor gave does not involve any exponential, such as x2.
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11/09/13

M K.

Okay, that would make sense. She did mention that it may not work out like that. Thank you!
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11/09/13

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