Inverse functions

Kim.j

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I need help on both of them please, ??
 

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These are not inverse functions. They are composite functions.

[MATH]f(b) = c \text { and } g(a) = b \implies f(g(a)) = f(b) = c.[/MATH]
[MATH]f(b) = c \text { and } g(c) = d \implies g(f(b)) = d.[/MATH]
That is all there is to composite functions.

And this is not a problem in pre-algebra.
 
These are not inverse functions. They are composite functions.

[MATH]f(b) = c \text { and } g(a) = b \implies f(g(a)) = f(b) = c.[/MATH]
[MATH]f(b) = c \text { and } g(b) = d \implies g(f(b)) = d.[/MATH]
That is all there is to composite functions.

And this is not a problem in pre-algebra.

Well they do turn out to be inverse functions because f(g(x))= g(f(x))= x.
 
@Subhotosh Khan

If you feel that my post is irredeemable, please delete it.

Otherwise, please change second line in LaTeX to

[MATH]f(b) = c \text { and } g(c) = d \implies g(f(b)) = g(c) = d.[/MATH] .....................................done

Perhaps (see @HallsofIvy ) my comment that this is a problem in function composition rather than inverse functions is misleading. It turns out that if the domain is restricted to non-negative numbers in this specific problem, f and g are inverses. Of course, if f and g are assumed to be inverses, the problem itself is trivial. It seems to me, however, illegitimate to deduce from a problem statement instructing ”Compute f(g(x)) and g(f(x))“ that f and g are inverses. Of course, f and g may be inverses; that is determined by taking their composition. If, however, you think I am being persnickety or confusing, you should of course delete the post or add this paragraph.
 
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Yes the function [MATH]f(x)=x^2+7, x \in \mathbb{R}[/MATH] doesn't have an inverse function, as it's not 1-1. The composite function [MATH]g(f(x))=|x|, x \in \mathbb{R}[/MATH]E.g. the function [MATH]f(x)=x^2+7, x≥0[/MATH] has an inverse function and [MATH]f(g(x))=x, x≥7[/MATH] and [MATH]g(f(x))=x, x≥0[/MATH]
 
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