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.

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 $n \le 1$.
