# Is the following function bijective?

Is this function bijective? Bijective means both onto and 1 to 1

$$F(x) = \frac{x^2+1}{x^2+2}$$

I'm not sure how to go about this.

Edit: The domain is ${\rm I\!R}$

• What is its domain? Jun 6, 2016 at 1:02
• Are we to assume that $F:\mathbb R\to \mathbb R$? Further, you could try sketching the graph to give you some idea.
– Em.
Jun 6, 2016 at 1:04
• To visualize the curve, it may be helpful to rewrite our expression as $1-\frac{1}{x^2+2}$. Jun 6, 2016 at 1:08
• It is an even function. It cannot be injective unless you are pruning its natural domain. Jun 6, 2016 at 1:09
• The domain is real numbers, sorry, I forgot to add that. I'll edit it. Jun 6, 2016 at 1:26

## 6 Answers

Assuming $F\colon \mathbb{R}\rightarrow \mathbb{R}$ as in @probablyme's comment.

Hint: draw a picture of the graph of the function. Can you find a horizontal line that intersects two points on the graph? If so, what can you say about the 1-to-1 property?

Alternative hint: does the function ever output negative numbers? If so, what can you say about the onto property?

Since $x$ is squared in all places where it occurs, the value of the function is always positive. So the function is not onto $\mathbb{R}$ and hence not bijective.

The function $F$:$\mathbb{R}\rightarrow \mathbb{R}$ is not a bijective function.

Because when $F(1)= 2/3, F(-1) = 2/3$, that invalidate the injection. (i.e. if $F(x) = F(y)$, $x$ equals to $y$) which shows that it is not a bijective function.

As par said, F(x) is not subjective on $\mathbb{R}$,for the range of F is the subset of $[0,+ \infty )$

However the function $F$:$\mathbb{[0,+\infty) }\rightarrow [0.5,1 )$ by checking the monotony and symmetry of function $F$.

Let f be a function from R+.Every function is onto. To show 1-1: Assume (x^2+1)/(x^2+2) =(y^2+1)/(y^2+2) . If x^2+1=y^2+1, x=y and we are done. Similarly if x^2+2=y^2+2. Assume x^2+1=alpha(y^2+1). Then necessarily x^2+2=alpha(y^2+2). But this is impossible.

• What do you mean every function is onto. If y =1, f (x) =1 is impossible. It's not onto. I don't understand the second half. Do you think you are proving it is 1-1 or not one-to-one one? Jun 6, 2016 at 5:04
• Every function is onto its range. It doesn't matter what I think i am proving. What did I prove? I assumed f(x)=f(y) and then if the numerator of f(x) is the numerator of f(y) then x=y, if the denominator of f(x) is the denominator of f(y) then x=y. Moreover they cannot be eual without eual numerator (with numerator and denominator being the obvious numerator and denominator shown in the problem) Jun 6, 2016 at 13:42
• Every function is onto its range but the range is not specified being onto refers to being onto the specified or implied codomain (otherwise the term is trivial). This function is onto [1/2, 1) but it is not onto R. It is not an onto real function because its image it not the real numbers. The stated domain was R, not R+. So if $x \ne - x; f(x) \ne f(-x)$. So it is neither 1-1 from R nor onto R so f:R->R is not a bijection but f:R+->image(f(R+)) is. (Actually f:R+U{0}->[1/2,1) is bijection.) I found it hard to follow your 1-1 argument and I was't sure if you were arguing it was or wasn't. Jun 6, 2016 at 15:52

Here is the graph of $f$.

Injectivity: $f(x) = f(y) \Rightarrow x = y$

This property is given, if every function value is not assigned to more than one ordinate $x$.

Which is not the case here, the graph of $F$ is symmetric to the $y$-axis, and for $x \ne 0$ we have $f(x) = f(-x)$.

Surjectivity: Every point of the codomain has a function value assigned.

The codomain is $\mathbb{R}$ (as nothing specific is said) and e.g. $y = 0$ is not in the image of $F$.

$f:\mathbb R \rightarrow \mathbb R$ is not a bijection as it is neither 1-1 nor onto.

However, $f:[0, \infty)\rightarrow [\frac 12,1)$ IS a bijection. Which only goes to show, that when you make a claim such as "f is bijection" it is essential you specify on what domain and range you are referring.

Claim 1: $f:\mathbb R \rightarrow \mathbb R$ is not 1-1.

Pf. Let $x\ne 0$. Then $x \ne -x$. But $x^2 = (-x)^2$ so $f(-x)=f (x)$.

Claim 2: $f:\mathbb R \rightarrow \mathbb R$ is not onto.

Pf: $x^2 + 1 > 0$ and $x^2 + 2 > 0$ so $f(x) >0$ so for any $y \le 0$ there are no $x$ so that $f(x) = y$.

Furthermore $x^2 + 2 > x^2 + 1$ so $f(x) < 1$ so for any $y \ge 1$ there are no $x$ so that $f(x) =y$.

Claim 3: $f: [0,x)\rightarrow [1/2,1)$ is 1-1.

Pf.
If $f(x) = f(y)$ then $\frac{x^2 + 1}{x^2 + 2} = \frac{y^2 + 1}{y^2 + 2}$.

So $(x^2 + 1)(y^2 + 2)=(x^2 +2)(y^2 + 1)$ so

$x^2y^2 + y^2 + 2x^2 + 2 = x^2y^2 + x^2 + 2y^2 + 2$ so

$x^2 = y^2$ so $\pm x = \pm y$. But as $x$ and $y$ are both in $[0,\infty)$ they are both non negative. So $x = y$. So is 1-1.

Claim 4: $f: [0,x)\rightarrow [1/2,1)$ is onto.

$f$ is clearly continuous. And $\lim_x\rightarrow \infty f(x) = 1$ clearly. $f(0) = 1/2$. So for all $y \in [1/2,1)$ the exists an $x \in [0,\infty)$ such that $f(x)$ by intermediate value theorem.

• Actually if $f:X \rightarrow Y$ is not onto you can always find a range that it is onto by restricting to the image. $f:X \rightarrow Image(f(X)) \subset Y$ will always be onto. Jun 6, 2016 at 2:29