Proving that a function doesn't intercept the $x$ axis

Question

We must prove that the function $f$ doesn't intercept the $x$ axis.

$$ f: \mathbb R \to \mathbb R , \quad f(x) = x^{2012} + 2x^2 + \frac{1}{2012} $$

This should mean that the function has no real roots, right? I tried to take the derivative of the function to find the function's critical point(s). I've found that the function's minimum point is at $f(0)$ which equals $\frac{1}{2012}$ (edited).

I would like to know how it is possible to solve such a problem without using calculus concepts and if we can show that $f(x)$ has no real roots. Thanks.

Answer 1

We can use the fact that any real number squared is non-negative; this means $f(x)=x^{2012}+2x^2+\frac{1}{2012}\geq0+0+\frac{1}{2012}>0$ for all real $x$. Since the function is positive for all real $x$, it cannot intercept the $x$ axis (equivalently, it has no real roots).

Answer 2

$$f(x)=0\implies x^{2012}+2x^2=-\frac1{2012}$$But $LHS$ is the sum of two things that are squared, so must be positive. $RHS$ is negative. So there can be no solutions.