Prove that a sequence is increasing (check the idea) 
4.5 A real sequence $\{x_n\}$ satisfies $7x_{n+1}=x_n^3+6$ for $n\ge1$. If $x_1=\frac{1}{2}$, prove that the sequence increases and find its limit. What happens if $x_1=\frac{3}{2}$ or if $x_1=\frac{5}{2}$?

In order to prove, that sequence is increasing, my idea is to take the first derivative of the function $y = (x^3 +6)/7$
Our result is $y' = 3x^2/7$ and this is positive. So our function is increasing function. Is this solution correct?
Are there other approaches that can prove the above statement without using derivatives?
Edited:
I agree, I was wrong. Here are my new arguments:
1) By induction I have proved that $0<x_n<1$.
2) $x_{n+1} - x_n = (x_n^3 + 6)/7 -x_n = x_n^3/7 - x_n + 6/7 $
But I don't know if we can conclude from this that $x_n^3/7 - x_n + 6/7 >0$
 A: No, not really.
You're basically correct that the function $y=(x^3+6)/7$ is increasing and the argument is basically correct. However that was not the real question. You were to prove that $x_n$ is an increasing sequence, which means that you have to prove that $x_{n+1} = (x_n^3+6)/7 \ge x_n$.
To solve the problem you should probably not use that approach. You should consider the inequality above which says that we should investigate the sign of $(x^3+6)/7-x$ or $x^3+6-7x$. The trick here is to factor the polynomial. We have $x^3-7x+6 = (x+3)(x-1)(x-2)$. We see that it can only change sign at $x=-3$, $x=1$ and $x=2$. So we have that:
$$x_{n+1} \begin{cases}\lt x_n & \text{if } 1 < x_n < 2\\
\lt x_n & \text{ if } x_n < -3 \\
\ge x_n & \text{ otherwise }\end{cases}$$ 
This is enough to see that the sequence will increase as long as $x_n<1$. But we see that $(x^3+6)/7<1$ if $0<x<1$ so the recursion formula will never pass or even reach $1$.
To prove that actually $\lim x_n=1$ we observe that $x_n$ is increasing and it's bounded so obviously $x_n$ is convergent, let say $L = \lim x_n$, then it's easy to see that $0 = \lim 7x_{n+1} - x_n^3 - 6 = 7L-L^3-6$. The solutions to that equation is only $L=-3$, $L=2$ and $L=1$, but only $L=1$ fits the requirements.
For $x_1=3/2$ we will with the similar reasoning have that $x_n$ decreases and it turns out that $x_n\to1$. And for $x_1=5/2$ it will increase and $x_n\to\infty$ (we observe that if having a limit it must be one of the above, but increasing from $5/2$ can't be one of them so it can't have a limit and therefor not bounded).
A: Given that $x_{n+1} = (x_{n}^3 + 6)/7$, when $x_1 = \frac{1}{2}$ we will find the limit in 3 steps:


*

*$\color{red}{0 < x_n  < 1 \text{ for all $n$.}}$ Note that if $0 < x_n < 1$ then $0 < x_n^3 + 6 < 7$, which in turn implies that $0 < x_{n+1} < 1$. So, starting with $x_1 = \frac{1}{2}$, we will get the entire sequence satisfying  the inequality $0 < x_n  < 1$.

*$\color{red}{x_n \text { is increasing. }}$ To show this, it is sufficient to show that if $0 <x_n < 1$ then $x_n < x_{n+1}$. To get this, we subtract $x_n$ from both sides of $x_{n+1} = (x_{n}^3 + 6)/7$ and get:
\begin{eqnarray*}x_{n+1} - x_{n} = \displaystyle\frac{x_{n}^3 + 6 - 7 x_n}{7} =  \displaystyle\frac{(x_n-1)(x_{n}^2 + x_n - 6)}{7} > 0 & \text{ when $0 < x_n < 1$} \end{eqnarray*}

*$\color{red}{x_n \text{ converges to 1}. }$ Since the sequence $(x_n)$ is increasing and bounded, it is convergent. Therefore, the limit $l \leq 1$ exists and satisfy: $l = (l^3 + 6)/7$. This gives us the limit as $l = 1$.


Similarly, when $x_1 = \frac{3}{2}$ we can again find it in 3 steps:


*

*$\color{red}{1 < x_n  < 2 \text{ for all $n$.}}$ If $1 < x_n < 2$ then $1 < x_n^3 + 6 < 14$, which in turn implies that $1 < x_{n+1} < 2$. So, starting with $x_1 = \frac{3}{2}$, we will get the entire sequence satisfying  the inequality $1 < x_n  < 2$.

*$\color{red}{x_n \text { is decreasing. }}$ To show this, it is sufficient to show that if $1 <x_n < 2$ then $x_n > x_{n+1}$. To get this, we subtract $x_n$ from both sides of $x_{n+1} = (x_{n}^3 + 6)/7$ and get:
\begin{eqnarray*}x_{n+1} - x_{n} = \displaystyle\frac{x_{n}^3 + 6 - 7 x_n}{7} =  \displaystyle\frac{(x_n-1)(x_{n}^2 + x_n - 6)}{7} < 0 & \text{ when $1 < x_n < 2$} \end{eqnarray*}

*$\color{red}{x_n \text{ converges to 1}. }$ Since the sequence $(x_n)$ is decreasing and bounded, it is convergent. Therefore, the limit $1 \leq l \leq 1.5$ exists and satisfy: $l = (l^3 + 6)/7$. This gives us the limit as $l = 1$.


Finally, when $x_{1} = \frac{5}{2}$ we can show that the sequence is unbounded above, and increasing. Therefore, limit does not exist on real line. This follows from the fact that
\begin{eqnarray*}x_{n+1} - x_{n} = \displaystyle\frac{x_{n}^3 + 6 - 7 x_n}{7} =  \displaystyle\frac{(x_n-1)(x_{n}^2 + x_n - 6)}{7} >0  & \text{ when $x_n > 2$} \end{eqnarray*}
Also, when $x_n > 2$, the function $\displaystyle\frac{(x_n-1)(x_{n}^2 + x_n - 6)}{7}$ is also increasing in $x_n$. This implies that the difference between $x_{n+1}$ and $x_{n}$ is always greater than or equal to the value:
$\displaystyle\frac{(x_1-1)(x_{1}^2 + x_1 - 6)}{7} = \frac{1.5(2.75)}{7}$. Therefore, the sequence $(x_n)$ is increasing and unbounded.

