Proof that $f(x)=4x^4-2x+1$ has no real roots. My thought was to: 1) hypothesis there are 2 real roots for this equation, 2) apply Rolle's theorem and come to a reductio ad absurdum and then if there aren't 2 real roots, it has to be 1. If there is 1 real root, this means that it has to have 3 non-real roots. But non real roots come in pairs, so either is 2 real- 2 non real, either 0 real- 4 non real. Therefore, it has no real roots. In case my thought is correct, the problem is that $f'(x)=0 => x^3=1/8$ doesn't lead me in reductio ad absurdum, because it has 1 real and 2 non- real roots. I'm stuck and I'm about to punch the desk. Please, release me from this martyrdom.

* $f'(x)=2(8x^3-1)$, so there's a single critical point: $\; x=\frac12$.
* $f''(x)=48x^2\ge 0$, so by the second derivative test, this critical point is a _minimum_ , and this minimum is an absolute minimum.
* $f(\frac12)=\frac14>0$.