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Definition of an injective function

$$ f(x_1) = f(x_2) => x_1 = x_2 $$

Well I have two functions that I can't prove...

$$ f(x) = x^3 + x $$ $$ f(x) = \frac{x}{1-log(x)} $$ log(x) denotes the logarithm base 10 function, well it doesn't matter for this problem, but whatever.

the first function $$ x_1^3 + x_1 = x_2^3 + x_2 $$ What can I do after this? Same with the second function, how can I manipulate in order to achieve $$ x_1 = x_2 $$

Nuno Mateus
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    Your first $f$ is increasing. – Angina Seng May 01 '19 at 19:32
  • and your second $f$ doesn't appear to be injective – J. W. Tanner May 01 '19 at 19:33
  • Hmmm, so If I prove that a function is monotonic increasing/decreasing) on its domain.... always increasing and is continuous then it's injective?

    Yes I saw later, the second isn't

     – Nuno Mateus May 01 '19 at 19:38
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    @NunoMateus: yes, if its strictly increasing or decreasing, it's injective – Vasili May 01 '19 at 19:39
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    What you have given are no functions at all. You are missing to tell what the domain and codomain is. You should give functions as $f: X\to Y$ with $x\mapsto f(x)$. That is espacially important when you are talking about properties like injective and surjective. – Cornman May 01 '19 at 19:52

3 Answers3

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For the first one: $$ x_1^3 + x_1 = x_2^3 + x_2 \rightarrow x_1^3-x_2^3+ x_1-x_2=0 \rightarrow (x_1 - x_2)(x_1^2+x_1 \cdot x_2+x_2^2+1)=0 \rightarrow x_1 - x_2=0$$ (the second term does not have real roots, discriminant is $x_2^2-4x_2^2-4<0)$

The second function is not injective.

Vasili
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I will assume your functions are defined on $\mathbb{R}$.

As noted in the comments, instead of using the definition of injective you have given, we can make use of the fact that the first function is strictly increasing.

This means, precisely, that

$$x > y \implies f(x) > f(y) \space \space \space (\forall x,y \in \mathbb{R})$$

To show that $f$ is increasing we can show that its derivative is always positive.

$\frac{df}{dx} = 3x^2 + 1$ which is clearly positive for all $x \in \mathbb{R}$.

Alternatively, without using calculus but making some smaller assumptions we can do this:

Assume $x>y$. Then, cubing both sides and noting that the cube function is itself strictly increasing, we have

$$x^3 > y^3$$

and then adding $x$ to one side and $y$ to the other (since $x>y$ this is valid)

$$x^3 + x > y^3 + y$$

$$f(x) > f(y)$$

and we are done.

Since $f$ is strictly increasing, it is injective.

雨が好きな人
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To provide an alternate way of showing injectivity for differentiable functions, every strictly increasing (or decreasing) function is injective (see here for a proof).

To show that a function is strictly increasing, show that its derivative is always positive. For example, for your first function, $f '(x) = 3x^2 + 1$, which is positive for any value of $x$. Therefore, the function is injective.

dxb
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