root of $g$ is smaller than that of $f$.

Question

For a fixed natural no. $n\ge4$, consider $$f(x)=x^3-(n+2)x^2+2nx-2,$$ $$g(x)=x^3-(n+3)x^2+2(n+1)x-2,$$ It seems that smallest root of $g$ is smaller than that of $f$. Can someone show how to prove it.

Note: All roots of $f$ and $g$ are positive.

Answer 1

I chose notation $f_n(x):=x^3-(n+2)x^2+2nx-2 \ $ (your $f(x)$).

In this way, $g(x) \equiv f_{n+1}(x)$.

(see curves ($C_n$) below for $n=4,5,6$ and $7$ in green, purple, orange and red resp. ; $x$ axis ranges from $0$ to $7.5$ ; I have used Desmos).

We observe that all curves ($C_n$) have in common two points $(0,-2)$ and $(2,-2)$ (this can be easily proved).

Let us remark that

$$f_{n+1}(x)=f_{n}(x)+\varphi(x) \ \ (*) \ \ \ \text{with} \ \ \varphi(x) := -x^2+2x = x(2-x).$$

Now, let us bracket the roots of $f_n$, using values $x=0,1,2,n,n+1$:

$$f_n(0)=-2 \ \ ; \ \ f_n(1)=n-3 \ \ ; \ \ f_n(2)=-2 \ \ ; \ \ f_n(n)=-2 \ \ ; \ \ f_n(n+1)=n^2-3.$$

Therefore, for all $n \geq 4$:

$$f_n(0)<0 \ \ ; \ \ f_n(1)>0 \ \ ; \ \ f_n(2)<0 \ \ ; \ \ f_n(n)<0 \ \ ; \ \ f_n(n+1)>0$$

Thus, as $f_n$ is continuous, for $n \geq 4$, $f_n$ possesses always three real roots :

• $\alpha_n \in (0,1)$

• $\beta_n \in (1,2)$

• $\gamma_n \in (n,n+1)$

It remains to show that $\alpha_{n+1} < \alpha_{n}$.

As $\varphi(x)>0$ on $(0,2)$, and $\alpha_n \in (0,1)$, we have $a_n:=\varphi(\alpha_n)>0$. Using relationship (*):

$$f_{n+1}(\alpha_n)=f_{n}(\alpha_n)+\varphi(\alpha_n)=0+a_n>0$$

Thus $f_{n+1}(x)$, being $<0$ for $x=0$ and $>0$ for $x=\alpha_n < 1$, and having a unique root in $(0,1)$, has this (leftmost) root $\alpha_{n+1}$ such that $\alpha_{n+1}<\alpha_{n}$.

Edit:

It is not uninteresting to remark that,

• $(\alpha_n)_n$ is a decreasing sequence converging to $0_+$,

• $(\beta_n)_n$ is an increasing sequence converging to $2_-$, and, as a consequence of Vieta's formulas,

• $|\gamma_n-n|\rightarrow 0$ as $n \rightarrow \infty$.