Why is my proof for “if $0 \leqslant x \leqslant 2$, then $-x^3 + 4x + 1 > 0$” is false?

Question

$$\text{if $0 \leqslant x \leqslant 2$, then $-x^3 + 4x + 1 > 0$}$$

$$x(4-x^2)>-1$$ $$x>\dfrac{1}{x^2-4}$$ if the last statement is smaller than $x$ then it is also smaller than $2$ because of first statement ($0 \leqslant x \leqslant 2$): $$\dfrac{1}{x^2-4}<2$$ $$1<2x^2-8$$ $$4.5<x^2$$ $$-2.121<x<2.121$$ Which contradicts the first statement because $x$ cannot be greater than $2$. My proof is wrong but why? (I took that question from Math for Computer Science book p. 12)

Answer 1

You made two mistakes and recovered: $$\dfrac{1}{x^2-4}<2 \not\Rightarrow 1<2(x^2-4) \ \ (\text{it must be} \ 1\color{red}>2(x^2-4), \text{because} \ x^2-4\le 0)\\ 4.5<x^2 \not\Rightarrow -2.121<x<2.121 \ (\text{it must be} \ 4.5\color{red}>x^2 \Rightarrow -2.121<x<2.121)$$

Alternatively, you can prove it as follows: $$-x^3 + 4x + 1 > 0 \iff x^3-4x-1<0 \iff x(x-2)(x+2)<1 \iff \\ x(x-2)(x+2)\le 0<1 \ \ \text{for} \ 0\le x\le 2 \ \text{(why?)}$$

Answer 2

You are correct that $-3/\sqrt2<x<3/\sqrt2$ because $x$ still lies between $0$ and $2$. You have found a wider bound for $x$, but it is not wrong. Note that the larger interval does not mean that $x$ takes all values between $\pm3/\sqrt2$.

An analogy could be that $x=3$, and we know that $2<x<4$. But it is also true that $-1<x<7$, or even $-\infty<x<\infty$.

Answer 3

Stating that a statement is smaller than a number makes no sense.

And from the fact that $\dfrac1{x^2-4}<2$ together with $0\leqslant x\leqslant2$, you can't deduce that $\dfrac1{x^2-4}<x$.

Finally, from $\dfrac1{x^2-4}<2$, what you deduce is that $1>x(x^2-4)$, since $x^2-4<0$.

Answer 4

1 + 4x - x$^3$ = 1 + x(4 - x$^2$).
For all x in [0,2], 0 < x, 0 < 4 - x$^2$.
Desired conclusion quickly follows.