Confusing Surjective Question

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May 6th 2010, 11:41 AM
craig

Confusing Surjective Question

Quote:

Let $A = \mathbb{R} \backslash \{ 4 \}$

Let $f(a) = \frac{2a + 3}{a-4}$ for all $a \in A$ .

Show $f$ is injective but not surjective.

Did this easily, injective was simple enough, and showed that $f \neq 4$

Quote:

For which element $b$ of $\mathbb{R}$ is it true that there's a bijective function $g: A \to \mathbb{R} \backslash \{ b \}$ such that $g(a) = \frac{2a + 3}{a-4}$ for all $a \in A$ .

No idea where to start with this, obviously need to show that this new function is now surjective as well as injective, but not sure how to do this seeing as I just proved that an extremely similar function isn't surjective!!

Any help would be greatly appreciated.
May 6th 2010, 12:07 PM
undefined

Quote:

Originally Posted by craig

Did this easily, injective was simple enough, and showed that $f \neq 4$

No idea where to start with this, obviously need to show that this new function is now surjective as well as injective, but not sure how to do this seeing as I just proved that an extremely similar function isn't surjective!!

Any help would be greatly appreciated.

We need the codomain to equal the image.

You can see that the limit of g(x) as x -> infinity is 2, and the limit of g(x) as x -> -infinity is also 2. In particular, g(x) never takes on the value of 2, and the image of g is R \ {2}.

So set that as the codomain and you're good to go.
May 6th 2010, 12:08 PM
Plato

Look at two cases: $b\not=2~\&~b\not=\frac{-3}{2}$