Prove that a function is injective

Question

I have the function $f(x)={{x^3+9}\over{x^2}}$ and I'm trying to prove that it is injective in $(-\infty, 0)$ through the definition.

What I've done:

$ f(a)=f(b)\Leftrightarrow {{a^3+9}\over{a^2}}={{b^3+9}\over{b^2}}\Leftrightarrow b^2(a^3+9)=a^2(b^3+9)\Leftrightarrow a^3b^2+9b^2=a^2b^3+9a^2\Leftrightarrow\\ a^3b^2-a^2b^3=9a^2-9b^2\Leftrightarrow a^2b^2(a-b)=9(a-b)(a+b)\Leftrightarrow a^2b^2=9(a+b) $

From this point on, I'm pretty much doing circles without ever reaching $f(a)=f(b)\Leftrightarrow a=b$.

Question:

How can I prove that $f(x)$ is an injective function? Any tip that will help me get unstuck from where I am and points to me the right direction will be greatly appreciated.

Answer 1

Note that to prove injectivity, you only need to show that if $a,b \in (-\infty,0)$ with $f(a)=f(b)$, then $a=b$.

There are two cases to consider:

• If $a=b$, then we are done.

• Otherwise, assume $a \neq b$. Continuing from where you left, we get $$a^2b^2=9(a+b)$$

Since $x^2 \geq 0$ for any $x \in \mathbb{R}$, it follows that $a^2b^2 \geq 0$.

On the other hand, as $a,b<0$, it follows that $9(a+b)<0$

So, we have arrived at a contradiction. Thus, this case is not possible.

This shows that our required function is injective.

Answer 2

Write $$ f(x)=x+\frac{9}{x^2} $$ and observe that $$ f'(x)=1-\frac{18}{x^3} $$ so that $f$ is increasing since $f'(x)>0$ for $x<0$ and thus injective.

Answer 3

Note that

$$f(x)={{x^3+9}\over{x^2}}=x+\frac{9}{x^2}\implies f'(x)=1-\frac{18}{x^3}>0$$

thus f is strictly increasing and injective in $(-\infty, 0)$ .