Real Analysis. Bounds, Infimum and Supremum

Question

Given that $S=\left\{x \mid 4x^2 > x^3 + x\right\}$.

(1) Determine whether $S$ is bounded.

(2) Determine their supremum and infimum.

I divided the equation by $x$ to have a quadratic. Then my roots are in decimals doesn't look correct.

Thank you!

Try and edit your question such to include attempts at a solution or a particular concept that you find hard to grasp, or it will get closed for not respecting the policy of the site. — Victor, Aug 01 '15 at 23:06
S is not bounded $$ x \to -\infty \ 4x^2> x^3+x\ +\infty > -\infty$$ — Khosrotash, Aug 01 '15 at 23:10
To solve the inequality ,be sure that "x>0 " then divide by "x" ,but if you are not sure : you must solve it both , 1st by x>0 ,second by x<0 — Khosrotash, Aug 01 '15 at 23:15
S={x|4x^2>x^3+x} include x=-1 ,-2, -3 ,-4.5 ,-6.56 ,-100 ,-10000,-10000000,...,-$\infty$ — Khosrotash, Aug 01 '15 at 23:17

score 1 · Answer 1 · answered Aug 01 '15 at 23:33

$$ 4x^2 > x^3 + x $$ Dividing both sides by $x$ and getting $4x > x^2 + 1$ is wrong. You can divide both sides by a positive number and do that, or by a negative number and get "$<$" instead of "$>$", but $x$ is sometimes positive and sometimes negative.

You have \begin{align} x^3 - 4x^2 + x < 0 \\[10pt] x(x^2 - 4x+1) < 0 \\[10pt] (x-0)\Big(x-(2-\sqrt 3)\Big)\Big(x-(2+\sqrt 3)\Big) < 0 \end{align}

Thus there are three places where the sign changes: $0$, $2-\sqrt 3$, and $2+\sqrt 3$. They divide the line into four intervals:

numbers less than $0$,
numbers between $0$ and $2-\sqrt3$,
numbers between $2-\sqrt3$ and $2+\sqrt 3$
numbers bigger than $2+\sqrt3$.
On the first interval, all three of the factors are negative so the product is negative.
On the second interval, the first factor is positive and the other two are negative, so the product is positive.
On the third interval, the first two factors are positive and the third is negative, so the product is negative.
On the fourth interval, all three factors are positive, so the product is positive.

So $S= (-\infty,0)\cup(2-\sqrt 3,\ 2+\sqrt 3)$, and from that you get the infimum and supremum.

Thanks so much. So S is Unbounded and the Infrimum is 2-sqrt(3) Supremum is 2+sqrt(3) — Soledolu Oluwatayo, Aug 01 '15 at 23:43
Then the Supremum is 0. So Unbounded sets can have Supremum and Infrimum? — Soledolu Oluwatayo, Aug 01 '15 at 23:53
The supremum is $2+\sqrt 3$. An unbounded set has a supremum and an infimum within the ordered set $[-\infty,+\infty]$, but lacks either a supremum or an infimum within the set $(-\infty,+\infty)$. If you think the supremum is $0$, then you're confused about something, but I can't say what unless I know how you reached that conclusion. ${}\qquad{}$ — Michael Hardy, Aug 02 '15 at 00:13

score 0 · Answer 2 · answered Aug 01 '15 at 23:13

0

1) $\forall x \in \mathbb{R} : x < 0 \Rightarrow x \in S$ (Why?).

So $S$ can't be bounded.

2) Can you figure it out yourself now?

answered Aug 01 '15 at 23:13

No I still don't get it. – Soledolu Oluwatayo Aug 01 '15 at 23:17
Be a little more specific please. Do you understand 1)? – Aug 01 '15 at 23:19
Did you try something like: Let $f(x)=4x^2-x^3+x$....? – Aug 01 '15 at 23:25
Do the numbers $-1,-2,-3,\cdots $ satisfy the inequality? – Matematleta Aug 01 '15 at 23:26
Yes they satisfy – Soledolu Oluwatayo Aug 01 '15 at 23:29
Does -10000 satisfy the inequality? – Aug 01 '15 at 23:31
Again, what exactly is it that you don't understand? – Aug 01 '15 at 23:40
1000 doesn't satisfy the inequality – Soledolu Oluwatayo Aug 01 '15 at 23:45
Sorry I think this is useless. Bye – Aug 01 '15 at 23:46

Real Analysis. Bounds, Infimum and Supremum

2 Answers2