*5 +1
x ----> 5x ------> 5x + 1 = f(x)
Reversing the steps gives:
-1 /5
5x + 1 -----> 5x ------> x
Applying this to x we get:
-1 /5
x ------> x - 1 ------> (x-1)/5 = f-1(x)
Brian
At 08:40 PM 2/2/97 EST, you wrote:
>A posting from G. Matthews talks about undoing operations to solve
>equations. I like this idea.
>
>In fact, it is one of the ways in which I introduce finding the
>formula for the inverse of certain functions.
>
>For example, given f(x) = 5x + 1, this says that to find the value of
>f(x), start with x, multiply by 5, then add 1 to the result.
>
>Since the inverse of a function "takes a value back from whence it
>came" we want to undo the above.
>
>To undo
> start with x, multiply by 5, then add 1 to the result,
>take the result, and
> subtract 1, then divide this by 5. (read the line above from
>right to left).
>Therefore the inverse function is
> f-1(x) = (x - 1)/5
>
>This works well for linear functions, and I don't try to go further
>with it. But it's another way to view the inverse of a function.
>
>Phil Mahler
>Middlesex CC
>Bedford, MA
>
______________________________________________________________________
Brian E. Smith
Dept of Mathematics TEL: 514-931-8731 Ext 1713
Dawson College FAX: 514-931-3567
3040 Sherbrooke St. W EMAIL:
Montreal (1) smithb@management.mcgill.ca
Quebec, Canada H3Z 1A4 (2) inbs@musicb.mcgill.ca
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It is not certain that everything is uncertain. Blaise Pascal (1670)
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