+1 vote
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Consider the following function

$f(x)=\frac{x}{2x+1} , \, x\not= -\frac{1}{2}$

Is the function a bijection?

Yes, this is a one-to-one function.

For onto, let's suppose function is invertible so $f^{-1}(x)=-\frac{x}{2x-1}\, x \not= \frac{1}{2}$

In the final answer, they have given that f if defined on $R \rightarrow R$ then it will be a bijection,but if my function has to be onto, then the co-domain must be $R-\frac{1}{2}$ and so if my function is defined on $R \rightarrow R-\{ \frac{1}{2} \}$, then it will be a bijection.

Please let me know what's correct?
recategorized | 59 views

A function is both one-to-one and onto if you can find a preimage of every element in the codomain.

Now, our domain is

$f(x)=\frac{x}{2x+1} , \, x\not= -\frac{1}{2}$

Now, say we get $y$

$y = \frac{x}{2x+1}$

$y(2x+1) = x$

$2xy + y - x = 0$

$x(2y-1) = -y$

$x = \frac{-y}{2y-1} , \, y\not= \frac{1}{2}$

Thus, if from the codomain of Real numbers, $\frac{1}{2}$ is excluded, for every element in codomain, there will exist a preimage.
0
So that means function will be a bijection for $R-\frac{1}{2} \rightarrow R-\frac{1}{2}$
+1
No.  $R-\{\frac{-1}{2}\} \rightarrow R-\{\frac{1}{2}\}$

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+1 vote