Polynomial inequality in real line

Question

Let $g (x) =x^6-x^5 +x^2-x +3, x $ takes all real values. Show that $g (x)>0$ for all $x $.

I differentiated and obtained further derivatives to understand the curvature of function. It turns out $g(x)-3$ is negative only in (0,1) .So I tried to find the minimum value of $ g(x)-3$ in (0,1). However, I haven't been able to find the minimum. I wonder if the approach I followed would work in this problem however, I would appreciate any better alternate method. Also is there any general approach to tackle such problems?

Answer 1

If $x<0$ or $>1$ then $g(x) >0+0+3$ since $x^{6} >x^{5}$ adn $x^{2} >x$ in this case. Now let $0\leq x \leq 1$. Then $g(x)=(x+x^{5})(x-1)+3 \geq (x+x^{5})(-1)+3\geq (1+1)(-1)+3 >0$.

Answer 2

$$x^6-x^5+x^2-x+3=x^6-x^5-x+1+x^2+2=$$ $$=(x-1)(x^5-1)+x^2+2\geq x^2+2>0.$$