2 3 X 2 4x

5 Answers v

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Hint: testify the contrapositive statement. If $x<0$, and so the polynomial on the left-hand side takes on a negative value at $x$.

answered Sep one, 2019 at 14:52

MalkounMalkoun

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If $x^five-4x^four+3x^3-x^2+3x-4 \geq 0 \Rightarrow 10(x^four+3x^two+iii) \geq (4x^4+x^2+4) \Rightarrow x \geq \frac{4x^iv+10^ii+4}{x^4+3x^2+iii}$. Since the expression on the right hand side is always possitive, information technology indeed shows $x\gt 0$.

answered Sep 1, 2019 at xv:52

L--Fifty--

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$$p(x) = x^five-(1+x+ten^ii)(4-7x+4x^2) $$ and both $1+x+x^2$ and $4-7x+4x^2$ have negative discriminants, so they always take positive values. If follows that for any $x<0$ we have $p(x)<x^5<0$, and so $p(x)\geq 0$ clearly implies $x\geq 0$.

answered Sep i, 2019 at 15:14

Jack D'AurizioJack D'Aurizio

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• $\begingroup$ Actually $ten\gt3$ and in that location are some values greater than $3$ for which $p(x)$ is still negative. $\endgroup$

  Sep one, 2019 at 15:23

• $\begingroup$ @Piquito: all correct, just the OP only needs to prove that $p(10)\geq 0$ implies $x\geq 0$. $\endgroup$

  Sep i, 2019 at xv:24

• $\begingroup$ Of course. Regards. $\endgroup$

  Sep ane, 2019 at 15:30

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COMMENT.-Equivalently yous take $$ten^five+3x^3+3x\ge4x^4+x^two+4\Rightarrow x\ge0$$ and information technology is clear that the quintic is greater than the quartic from a certain positive value of $x$ (this value is a little more $3$.

answered Sep 1, 2019 at 15:28

PiquitoPiquito

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• $\begingroup$ I am deplorable but finding in a rigorous way this value "slightly" greater than $iii$ is as difficult every bit the initial trouble. $\endgroup$

  Sep one, 2019 at 16:30

• $\begingroup$ No, when you ask that $x$ be positive. Thanks you lot. $\endgroup$

  Sep two, 2019 at xviii:59

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Yous have that $$ \begin{gathered} p(10) = x^5 - 4x^4 + 3x^3 - x^2 + 3x - iv = \hfill \\ \hfill \\ = x^three \left( {ten^2 - 4x + iii} \correct) - \left( {x^2 - 3x + 4} \correct) \hfill \\ \cease{gathered} $$ At present:

1. $$-\left( {10^2 - 3x + four} \correct)<0 \;\;\forall x\;\; \in \mathbb R$$
2. $$ {x^2 - 4x + 3}>0\;\; \forall x\;\;<0$$
3. $$x^3<0 \forall x\;\;<0$$ Therefore, for every x<0 you have that p(x)<0. Hence, if p(x)=0 it must be ten>0

In lodge to prove that if p(x)=0 and then ten>3 we have that

1. If $$ 0< x <1 $$ then $$ f(x) > 0 $$

2. If $$ 1< x <3 $$ and then $$ f(x) < 0 $$

3. If $$ten=0, ten=1,x=3$$ then $$f(x)=0$$

4. f has a local maximum at $$ x_0 = \frac{1} {five}\left( {viii - \sqrt {19} } \right) $$ and $$ f(x_0 ) = \frac{{2\left( {2147\sqrt {19} - 8986} \right)}} {{3125}} < \frac{1} {2} $$

5. $$chiliad(x)=-(x^two-3x+4) \leq -\frac{7}{4} \forall x \in \mathbb R$$

Therefore if $$ 0 \leqslant x \leqslant 3 $$ it is $$ p(10) = f(x) + g\left( ten \right) \leqslant \frac{i} {two} - \frac{7} {iv} < 0 $$ It follows that if p(x)=0 and then x>iii

answered Sep 1, 2019 at eighteen:14

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• $\begingroup$ I got a mistake while I'm writing. Instead -(x^2-3x+iv)<0 for all x I wrote (x^2-4x+3)<0!!! Distressing $\endgroup$

  Sep 1, 2019 at xviii:29

2 3 X 2 4x,

Source: https://math.stackexchange.com/questions/3341045/proof-if-x5-4x43x3-x23x-4%E2%89%A50-then-x%E2%89%A50

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