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Re: Algebra 2/Trig Regents
Posted:
Jun 23, 2011 7:21 AM
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Yes Nick I agree with you. When the authors of texts are ambiguous then
how can we be so strict with the grading? Accepting all answers was the
right decision. If they had restricted the domain to begin with, the
problem would be moot.
----- Original Message -----
From: "Nick B" <nbianculli@gmail.com>
To: <nyshsmath@mathforum.org>
Sent: Wednesday, June 22, 2011 8:46 PM
Subject: Re: Algebra 2/Trig Regents
>I have a few precalc books lying around in my office...here are some of the
>offerings:
>
> Precalculus by Barnett, Ziegler, Byleen 6th Ed
>
> Definition of an inverse function:
>
> If f is a one-to-one function, then the inverse of f, denoted f^-1, is the
> function formed by reversing all the ordered points in f. Thus,
>
> f^-1 = {(y, x) | (x, y) is in f}
>
> If f is not one-to-one, then f does not have an inverse and f^-1 does not
> exist.
>
> Blitzer "Precalculus Essentials" 2nd Edition (I used this book when I
> taught Precalculus at Stony Brook University a few years ago):
>
> Definition of an Inverse Function:
>
> Let f and g be two functions such that
>
> f(g(x)) = x for every x in the domain of g
>
> and
>
> g(f(x)) = x for every x in the domain of f
>
> The function g is the inverse of the function of f and is denoted by f^-1
> (read "f-inverse"). Thus f(f^-1(x)) = x and f^-1(f(x)) = x. The domain
> of f is equal to the range of f^-1 and vice versa.
>
> Precalculus: Graphs and Models 3rd Ed by Bittinger, Beecher, Ellenbogen,
> and Penna Page 350
>
> If the inverse of f is also a function, it is named f^-1 (read
> "f-inverse")
>
> Functions Modeling Change by Connally, Hughes-Hallett, Gleason etc. (these
> are the same authors of the Reform Calculus books):
>
> Definiton of an Inverse Function:
>
> Suppose Q = f(t) is a function with the property that each value Q
> determines exactly one value of t. Then f has an inverse function, f^-1
> and
>
> f^-1(Q) = t if and only if Q = f(t)
>
> If a function has an inverse, it is said to be invertible
>
> This book goes on to talk about y = sin^-1(x) and y = cos^-1(x) as inverse
> functions.
>
> I think you could probably find a few books that allow f^-1(x) to be a
> relation and not a function, but I think the vast majority would not take
> this approach. Then again, it's not a battle of how many textbooks one can
> find to back up his or her conventions. It seems as though there are two
> ways to approach the notation f^-1, so I think SED got it right in the
> end.
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