Confusing Surjective Question

**craig** · May 6th 2010, 11:41 AM

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 $f \neq 4$

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.

**undefined** · May 6th 2010, 12:07 PM

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.

**Plato** · May 6th 2010, 12:08 PM

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

Thread: Confusing Surjective Question

LinkBack

Thread Tools

Display

Confusing Surjective Question

Similar Math Help Forum Discussions

Question on a surjective map

Injective, surjective, inclusion and exclusion question

prove that if g o f is surjective then g is surjective

Confusing Question

Confusing Question

Search Tags