Prove or disprove: $\sup\left \{ x\in\mathbb{R}\mid x^2-5x+6\leq0 \right \}=3$

Question

No homework:http://www2.mathematik.hu-berlin.de/~gaggle/S15/MATHINFO/UEBUNG/nachholklausur.pdf

Prove or disprove: $\sup\left \{ x\in\mathbb{R}\mid x^2-5x+6\leq0 \right\} =3$

I would say the statement is true because if we use the p-q-formula:

$$x_{1,2}= \frac{5}{2}\pm\sqrt{\left(\frac{-5}{2}\right)^2-6}$$

$$x_{1,2}= 2.5\pm\sqrt{\frac{25}{4}-6}$$

$$x_{1,2}= 2.5\pm\sqrt{\frac{1}{4}}$$

$$x_{1,2}= 2.5\pm\left(\frac{1}{2}\right)$$

$$x_1=3$$

$$x_2=2$$

Actually, I don't really understand the statement, what is it saying? That the supremum of the function is 3?

Then statement is wrong because we got as second solution 2 which is smaller than 3, so the supremum is 2 and not 3..?

Answer 1

The statement to prove or disprove is, reformulated, equivalent to the following two conditions

• if $x$ is a real solution to $x^2-5x +6\leq 0$, then $x\leq 3$;

• if $x>3$, then is not a solution to $x^2-5x +6$.

The statement is true: solving $x^2-5x +6 = 0$, we see that the only two solutions are $2$ and $3$. As the polynomial function is negative between the root, we have $x^2-5x +6 \leq 0$ if, and only if, $x\in[2,3]$.

Here is the graph of the polynomial function $f$ defined by $f(x)=x^2-5x +6$. The set $S\stackrel{\rm def}{=} \{x: f(x) \leq 0\}$ is the part of the $x$-axis for which the curve is below $0$. Since $S=[2,3]$, we have $\sup S = \sup [2,3] = 3$.

Answer 2

I think I know what to say next. I am pasting in a graph of $y = x^3 - 3x.$ What is $$\sup\left \{ x\in\mathbb{R}|x^{3}-3x\leq0 \right \}?$$