# Inverse functions

#### Kim.j

##### New member
I need help on both of them please,

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#### Subhotosh Khan

##### Super Moderator
Staff member
I need help on both of them please,
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

#### JeffM

##### Elite Member
These are not inverse functions. They are composite functions.

$$\displaystyle f(b) = c \text { and } g(a) = b \implies f(g(a)) = f(b) = c.$$

$$\displaystyle f(b) = c \text { and } g(c) = d \implies g(f(b)) = d.$$

That is all there is to composite functions.

And this is not a problem in pre-algebra.

#### HallsofIvy

##### Elite Member
These are not inverse functions. They are composite functions.

$$\displaystyle f(b) = c \text { and } g(a) = b \implies f(g(a)) = f(b) = c.$$

$$\displaystyle f(b) = c \text { and } g(b) = d \implies g(f(b)) = d.$$

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.

#### JeffM

##### Elite Member
@Subhotosh Khan

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

Otherwise, please change second line in LaTeX to

$$\displaystyle f(b) = c \text { and } g(c) = d \implies g(f(b)) = g(c) = d.$$ .....................................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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lex

#### lex

##### Full Member
Yes the function $$\displaystyle f(x)=x^2+7, x \in \mathbb{R}$$ doesn't have an inverse function, as it's not 1-1. The composite function $$\displaystyle g(f(x))=|x|, x \in \mathbb{R}$$
E.g. the function $$\displaystyle f(x)=x^2+7, x≥0$$ has an inverse function and $$\displaystyle f(g(x))=x, x≥7$$ and $$\displaystyle g(f(x))=x, x≥0$$