Number of positive integers $n$ The question is to find out the number of positive integers $n$ such that $nx^4+4x+3 \leq 0 $ for some real $x$ (without using a graphic calculator).
My attempt at the solution:
We have $nx^4+4x+3 \leq 0 $ which is $nx^4 \leq -3-4x$ or $n \leq -\frac{(3+4x)}{x^4}$. It can easily be observed that for $x=-1$ the condition of $n$ being a positive integer is satisfied. But I could not get others. Moreover to satisfy that $n$ is positive we must have $x<-3/4$.
Please help me in this regard. Thanks.
 A: Solution WITHOUT calculus.
Note that
$$nx^4+4x+3=(n-1)x^4+x^4+4x+3=(n-1)x^4+((x-1)^2+2)(x+1)^2.$$ 
Hence for $n=1$, we have that the above polynomial is $0$ at $x=-1$.
If $n\geq 2$ then, for any $x\in\mathbb{R}$,
$$nx^4+4x+3\geq x^4+2(x+1)^2>0$$
because the squares $x^4$ and $(x+1)^2$ are zero for different values of $x$ (namely $x=0$ and $x=-1$).
A: Differentiate to get $4 n x^{3} + 4$. This has the single root $x_{0} = - \dfrac{1}{\sqrt[3]{n}}$ which is a minimum for $f(x) = nx^4+4x+3$. We want $f(x_{0}) \le 0$, or
$$
0 \ge n x_{0}^{4} + 4 x_{0} + 3 = n (-\dfrac{1}{n}) x_{0} + 4 x_{0} + 3 = 3 (x_{0} + 1).
$$
Thus we must have 
$$
- \dfrac{1}{\sqrt[3]{n}}= x_{0} \le -1,
$$
or $n \le 1$.
A: Let $f_n(x)=nx^4+4x+3$ $\quad$ ($n \in \mathbb N$).
Then $f_n'(x)=0$ iff $x=-\frac{1}{n^{1/3}}$ and $f_n''(-\frac{1}{n^{1/3}}) >0$.
Hence, since $f_n(x) \to \infty$  for $x \to \pm \infty$, $f_n$ has an absolut minimum at $-\frac{1}{n^{1/3}}$.
It is easy to see that $f_n(-\frac{1}{n^{1/3}})  \ge 0$ and that $f_n(-\frac{1}{n^{1/3}})  > 0$ if $n \ge 2$.
Thus: only for $n=1$ there is $x$ with $f_n(x) \le 0$: $x=-1$
A: $x = 0$ cannot be a root so suppose $x > 0$.
The equation is equivalently written as $nx^3+3/x = -4$. 
But by AM GM, $nx^3+3/x \ge 4 \sqrt[4]{n} > -4$ so no positive roots.
Likewise, let $x = -r$ where $r$ is a positive number.
Then we have $nr^4-4r+3 = 0$ or $nr^3+3/r = 4$.
Again by AM GM, we have $nr^3+3/r \ge 4\sqrt[4]{n} \ge 4$ with equality at $n = 1$ 
which is the unique positive integer.
