How many zeros does $f(x)= 3x^4 + x + 2 $ have?

Question

How many zeros does this function have? $$f(x)= 3x^4 + x + 2 $$

Answer 1

Hint:

$$ f(x)= 3x^4 + x + 2$$ for $x\in \mathbb R$ $$f'(x)= 12x^3 +1=0\implies x=-\frac{1}{\sqrt[3]{12}}$$

$$f\left(-\frac{1}{\sqrt[3]{12}}\right)=2-\frac{1}{4}\left(\frac32\right)^{2/3}\ge0$$

Answer 2

For $x\ge -2$, we have $(x+2)\ge 0$ and $3x^4>0$, so $f(x)>0$.

For $x<-2$, we have $3x^4+x=x(3x^3+1)=(-x)(-1-3x^3)>(2)(-1+3\cdot 2^3)=46$, so $f(x)>0$.

Answer 3

conjecturing it has no real roots is easy at first glance it can be seen the function is never very small.

So you can try to look for the minimum by differentiating:

$f'(x)=12x^3+1$. So the critical points are at $12x^3=-1 \iff x^3=\frac{1}{12} \iff x=-\sqrt[3]\frac{1}{12}$.

$f(-\sqrt[3]\frac{1}{12})=p-s+2$ where the first term is positive and the second one is really close to zero so the result is positive.

Answer 4

$f(x)$ has $4$ zeros by the fundamental theorem of algebra. Here is an alternative way to find that none of them is real:

$$ \ \ \ \ \ \ \ \ 3x^4+x+2=0 \ \ \ \ \ (1)\\ 3x^4=-x-2,$$ which, as $0\ne-2$, is equivalent to $$3x^3=-1-\frac{2}{x}.$$ Letting $g(x)=3x^3$ and $\displaystyle h(x)=-1-\frac{2}{x}$ we have $$ \left\{ \begin{array}{c} x>0 \\ g(x)=h(x)\\ g(x)>0 \\ h(x)<-1 \\ \end{array} \right. \text{and} \left\{ \begin{array}{c} x<0 \\ g(x)=h(x)\\g(x)<0 \\ h(x)>-1 \\ \end{array} \right. .$$

The first has no solutions whereas the second implies $$\left\{ \begin{array}{c} -1<g(x)<0 \\ -1<h(x)<0 \\ \end{array} \right. \leftrightarrow \left\{ \begin{array}{c} -\frac{1}{\sqrt[3]{3}}<x<0 \\ x<-2 \\ \end{array} \right. ,$$ which is an inconsistent system. Thus, $(1)$ is impossible in $\mathbb{R}$.