Checking if a function is one-one(injective)

Question

So I have this function:

$\dfrac{x^2-2}{x^2+2x}$ and im supposed to find if it is one-one.

I've tried taking out the drivative of it and check if its $>0$ or $<0$ but im facing trouble there.

I've also heard that you can do it with a graph but I'm looking for a more algebraic approach than a geometric one?

Isn't there a sureshot way of finding out if a function is one one or not?

Answer 1

Consider, for example, the equation $f(x)=2$. This gives $x^{2}-2=2(x^{2}+2x)$ or $x^{2}+4x+2=0$. There are two solutions of this quadratic: $x =-2\pm \sqrt 2$. Hence $f$ is not injective. In fact, every value in the range of $f$ except $1$ is attained at exactly two points whereas the value $1$ is attained only at $x=-1$.

Answer 2

This is all false. The if-statement fails. Kavi Rama Murthy's answer is correct.

Such a function would be injective if and only if it was strictly monotone in its domain. When we take $f'$, we get $$\frac{2x(x^2+2x) - (2x+2)(x^2-2)}{(x^2+2x)^2} = \frac{2x^2 + 4x + 4}{(x^2+2x)^2}$$ Determine domain of $f$ and see if it's either strictly increasing or strictly decreasing in the whole domain by the above.

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Here's my failure. Theorem. Let $f:D\to\mathbb R$, where $D\subseteq \mathbb R$ is connected. Then $f$ is injective if $f'>0$ or $f'<0$ (in $D$).

The domain for the function in question is not connected, hence the theorem is inapplicable.

Answer 3

Whether a function is injective (one-one) essentially requires a domain $A$ and r ange $B$ like $f:A \to B$., without this specification the question does not arise. The natural domain of $f(x)=\frac{x^2-2}{x(x+2)}$ is $\Re -\{0, -2\}$. Next, monotonicity of $f(x)$ is decided if is $f'(x)$ is positive-definite or negative-definite is decide by $f'(x)>0 or <0.$ So here $$ f'(x)=\frac{2[(x+1)^2+1]}{x^2(x+2)^2}>0.$$ So it is a increasing function in three largest domains: $$D_1=(-\infty, -2),~ D_2=(-2,0), ~ D_3 =(0,\infty).$$ Hence it is injective in tjese three largest domains. See $f(x)$ in the Fig. below:

Answer 4

I have a much simpler method to check whether a function is one-one or not.

As we know a one-one function can be defined as function from A to B such that each element of Set A has a unique element on Set B.

Whenever f(x)=f(y), then x=y

And equivalently, if x ≠ y, then f(x) ≠ f(y)

Formally, it is stated as, if f(x)=f(y) implies x=y, then f is one-to-one mapped or f is 1-1.

Now we take two elements $x_1$ and $x_2$ and assume that $f(x_1) = f(x_2)$ and ultimately prove that $x_1 = x_2$ only then can we prove that the given function is one-one.

You may refer this - How to determine if a function is one-to-one?