Vivian L. answered • 11/09/13
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Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACH
Hi M;
This is all I can answer...
f(x)=√(x-3)
As you explained, domain is 3 to positive infinity because a negative number cannot be square-rooted.
As you explained, range is 0 to positive infinity.
f-1=x2+3
The domain of the inverse is negative infinity to positive infinity, which is not a mirror image of the range of the original equation, as you explain.
However, the absolute value of the domain is 0 to positive infinity because...
Anything squared is a positive number or zero, as you already know.
And, as you already know, the minimum value of the range is...
f-1=02+3
f-1=3
The range of the inverse is 3 to positive infinity, a mirror of the domain of the original equation, as you already know.
That is all I can contribute. I hope it helps.
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
M K.
11/09/13