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I want to prove that $h(x)=x^3 +2x+1$ is a $1-1$ function to show that it is invertible on all of $\mathbb{R}$.

This my attempt: Let $x_1,x_2\in \mathbb{R}$ where $x_1\neq x_2$.

Suppose for contradiction $h(x_1)=h(x_2)$.

  • Then $h(x_1)=x_1^3 +2x_1+1$ and $h(x_2)=x_2^3 +2x_2+1$.
  • $x_1^3 +2x=x_2^3 +2x_2$

Then where do I go from here to show that $x_1=x_2$?

2 Answers2

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An easier way: The derivative is $$h'(x) = 3x^2 + 2$$ is strictly positive for all $x$, and so the function is strictly increasing. Formal justification could be made by, say, Rolle's theorem.

Alternatively, given $$x_1^3 + 2x_1 = x_2^3 + 2x_2$$

this can be rearranged as

$$0 = x_1^3 - x_2^3 + 2x_1 - 2x_2 = (x_1 - x_2)(x_1^2 + x_1 x_2 + x_2^2) + 2(x_1 - x_2)$$

Dividing by the non-zero quantity $x_1 - x_2$, this shows that

$$0 = x_1^2 + x_1 x_2 + x_2^2 + 2$$

This is a quadratic equation in $x_1$, whose discriminant is

$$x_2^2 - 4 (x_2^2 + 2)(1) = -3x_2^2 - 8$$

As this is strictly negative, there are no real solutions.

  • Thanks. I think it's better if I use the easier method. Also, can I informally just plot the function $h(x)$ and show that it passes the horizontal test to be invertible? –  Nov 06 '13 at 05:30
  • @AJR The informal plot doesn't really constitute a proof for most people; it's good to have the intuition and picture, but it's not very formal. –  Nov 06 '13 at 05:32
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$$x^3+2x=y^3+2y\iff (x^3-y^3)+2(x-y)=0\iff$$

$$(x-y)\left[x^2+xy+y^2+2\right]=0$$

To prove the part inside the square parentheses cannot be zero consider it as a quadratic in $\;x\;$ whose discriminant is

$$\Delta=y^2-4(y^2+2)=-3y^2-8<0\;\;\;\forall\;y$$

DonAntonio
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