Prove function is < 0 
For all $x$ if $x^6 + 3x^4 - 3x < 0$ then $0 < x < 1$. Prove this.
(1) Find the negation
(2) Prove

(1) The negation is simply, $\exists x$, $x^6 + 3x^4 - 3x < 0 \wedge (x \le 0 \vee x \ge 1)$
(2) The proof is the difficult part here.
We prove the contrapositive. It is easy to prove it for the condition that $x \le 0$ but it is harder for $x \ge 1$.
 A: The first derivative of $f(x)=x^6+3x^4-3x$ is $f'(x)=6x^5+12x^3 -3.$ Checking $f'(x),$ we see that if $x>1$, $ f'(x)>0$ ($f'(1)=15$, and 6x^5 + 12x^3 increase as you increase $x$ beyond $1$). Now, $f(1)= 1,$ and since $f'(x)>0$ for all $x \geq 1,$ $f(x)$ must be larger than $1$ for each value of $x$ greater than $1.$ 
If $x<0,$ notice similarly that $f'(x)<0$ (odd powers of negative numbers are always negative, and then you subtract 3 to boot). This indicates $f(x)$ is decreasing on the interval $(-\infty, 0].$ Since $f(0)=0$ (one root of the function right there!) we have that $f(x)>0$ for each value of $x$ less than $0.$
Therefore, it is impossible that $f(x)$ has a zero if $x$ is outside of the interval $[0,1].$ Hence, all zeros of $f(x)$ must occur between $0\leq x \leq 1.$ 
A: If $x <-1$, then $f (x)>f (-x)$ so it will suffice to check the cases for $x>1$. Notice that $x^6>x,$ since $(x\cdot x\dots x>x\cdot 1\dots 1=x $. Can you finish with the same argument for $3x^4$ and $3x $? What can you conclude for values of $f(x) $ outside of the unit interval?
A: For the second portion, set 
$$
x(x^5+3x^3-3)=0
$$
Which yields $x=0$ or $x^5+3x^3-3$. If $x=0$, then the strict inequality does not hold and clearly if $x<0$, the product is positive.
Furthermore, checking the derivative of  $x^5+3x^3-3$, we see that the function is strictly increasing, hence the zero of the odd function is unique. Then, that the zero occurs in between $0$ and $1$ follows form the intermediate value theorem. Again, you can eliminate options larger than 1 since the function is increasing.
