Proving injectivity of a certain polynomial function

Question

So I have to find the second derivative of the inverse function of:

$$f_{(x)}=3x^4+2x^3-8x^2-20x-160 ; x\ge 2$$

which I already have done, but I still need to prove the function is injective over the given domain.

I've tried the classic $f_{(x_1)}=f_{(x_2)}~$, but I end up with a factor I'm unable to simplify by using elementary methods.

$$f_{(x_1)}=f_{(x_2)}~$$ $$3x_1^4+2x_1^3-8x_1^2-20x_1-160=3x_2^4+2x_2^3-8x_2^2-20x_2-160$$ $$3(x_1^4-x_2^4)+2(x_1^3-x_2^3)-8(x_1^2-x_2^2)-20(x_1-x_2)=0$$ $$3(x_1^2+x_2^2)(x_1+x_2)(x_1-x_2)+2(x_1-x_2)(x_1^2+x_1x_2+x_2^2)-8(x_1+x_2)(x_1-x_2)-20(x_1-x_2)=0$$ $$(x_1-x_2)(3(x_1^2+x_2^2)(x_1+x_2)+2(x_1^2+x_1x_2+x_2^2)-8(x_1+x_2)-20)=0$$

so I found the $(x_1-x_2)$ factor I needed, but I can't find a way to simplify or to factor the second expression, or a way to prove it can't be equal to 0 over the function's domain.

How would one do it? Or is another way of proving the function's injectivity, other than graphing, recommended? If someone could prove it and tell me the procedure they used, it would be much appreciated.

Answer 1

For $x\geq2$ we obtain: $$(3x^4+2x^3-8x^2-20x-160)'=2(6x^3+3x^2-8x-10)=$$ $$=2(6x^3-12x^2+15x^2-30x+22x-44+34)=$$ $$=2(x-2)(6x^2+15x+22)+68>0,$$ which gives which you want.

Answer 2

$f'(x) = 12x^3+6x^2-16x-20 = x(12x^2+6x-16)-20$. Then $f'(2)=68>0$ and $x(12x^2+6x-16)>20$ for all $x \geq 2$ so $f'(x)>0$ for all $x \geq 2$. Then $f$ is strictly increasing over this domain, so it is injective.

Answer 3

One could also continue from $f(x_1)-f(x_2)= (x_1-x_2)(3(x_1^2+x_2^2)(x_1+x_2)+2(x_1^2+x_1x_2+x_2^2)-8(x_1+x_2)-20)=0$ at the end of the original question. The desired result $x_1=x_2$ would follows from $3(x_1^2+x_2^2)(x_1+x_2)+2(x_1^2+x_1x_2+x_2^2)-8(x_1+x_2)-20 \not=0$ for $x_1,x_2\geq2$.
Note that $3(x_1^2+x_2^2)(x_1+x_2)-8(x_1+x_2)= (x_1+x_2)[3(x_1^2+x_2^2)-8]\geq 4\cdot 16=64$. Accordingly $3(x_1^2+x_2^2)(x_1+x_2)+2(x_1^2+x_1x_2+x_2^2)-8(x_1+x_2)-20\geq64+24-20>0$ as desired.