Find the root for the function $2+x^{2}-x^{3}$

Question

How do we prove that the root for $$2+x^{2}-x^{3}$$ exits in the interval $[2,3]$??

no it isn't, we will get 2 by inserting -1 – onelessproblem Jun 10 '17 at 11:09 — onelessproblem, Jun 10 '17 at 11:09
are you sure?$$ 2+1+1\neq 0$$ – Dr. Sonnhard Graubner Jun 10 '17 at 11:09 — Dr. Sonnhard Graubner, Jun 10 '17 at 11:09
You don't prove that, since it isn' true. – Harald Hanche-Olsen Jun 10 '17 at 11:13 — Harald Hanche-Olsen, Jun 10 '17 at 11:13
You cannot because it is not true. – Emilio Novati Jun 10 '17 at 11:17 — Emilio Novati, Jun 10 '17 at 11:17
@onelessproblem I advise you to delete this question. – Jean Marie Jun 10 '17 at 11:19 — Jean Marie, Jun 10 '17 at 11:19
Why? is it too basic for this community? – onelessproblem Jun 10 '17 at 11:20 — onelessproblem, Jun 10 '17 at 11:20

score 3 · Answer 1 · answered Jun 10 '17 at 11:14

There is no root in the interval $(2,3)$.

To argue there is a root in $(1,2)$:

Polynomials are continuous.
$f(1)=2 >0$
$f(2)=-2 < 0$

Therefore, the function must have at least one $0$ value in the interval.

The root lone real root is $f\left( x_{0} \right) = 0$ with $$ x_{0} = \frac{1}{3} \left(\sqrt[3]{28 + 3 \sqrt{87}}+\sqrt[3]{28-3 \sqrt{87}}+1\right) \approx 1.69562 $$

score 0 · Answer 2 · answered Jun 10 '17 at 11:16

0

Let $f(x)=2+x^2-x^3$. Since $$f'(x)=2x-3x^2=(2-3x)x<0$$ when $x\in (2,3)$, $$f(x)<f(2)=-2$$ for any $x\in (2,3)$.

So there is no root.

answered Jun 10 '17 at 11:16

Paul

20,553

Find the root for the function $2+x^{2}-x^{3}$

2 Answers2