I tried several coefficients before $x^4$ and $x$, but I didn't get it unless adding a constant term.
Asked
Active
Viewed 69 times
3
-
May be your problem means a x^4 + b x or simply x^4 + a x – Claude Leibovici Dec 09 '13 at 07:13
-
What do u mean by always above? What can we do when $x\to +\infty$? – Mikasa Dec 09 '13 at 07:13
-
I mean for every $x$, the polynomial's value at that $x$ is greater than $x^3$. – user114515 Dec 09 '13 at 07:15
5 Answers
2
No, since in the interval $[0,1]$, the $x$ term will be the dominant one, we know that for the hypothesis to be true, $nx>1$, $n \in R$.Hence, We need $x>1/n$ $\forall x \in [0,1]$ and, we can't have that for all $x \in [0,1]$ because any positive number $'k'$, we will have $\infty$ positive numbers in $ [0,1]$ such that they are $ < k$.
Or
To Prove:
$ax^4-x^3+cx>=0$ for all $x \in R$,now we know that at $x=0$ the function is 0 and since the derivative of the function at $0$ isn't $0$, it won't have an inflexion point, hence will change sign at $x=0$, hence proved.
Isomorphic
- 1,182
1
If your function is only in terms of $x^4$ and $x$, then the answer is no, because your function will equal $0$ at $x=0$, meaning that it will coincide with $y = x^3$ there.
John
- 26,319
1
$ax^4+bx\geqslant x^3\iff$ $P(x)=ax^4-x^3+bx\geqslant0$. But for $x=0$ we already have $P(x)=0$, and since $P(x)$ is not allowed to be negative, this means that $x=0$ must be a minimum of $P(x)$. But local extremes are to be found among the roots of the derivative, which in our case is $4ax^3$$-3x^2$$+b$ which does not allow $x=0$ as a root, unless b itself is also $0$, in which case the term in x vanishes, thus breaking the rules laid out in the enunciation of the problem. If however b is allowed to be $0$, then the other two roots are both $\displaystyle\frac3{4a}$ , in which case we must have $P\left(\frac3{4a}\right)\geqslant0$, and all that's left is to solve for a, with the mention that when multiplying both sides of an inequality with a negative number, the symbol must be reversed, so there will have to be a discussion as to the sign of a.
Lucian
- 48,334
- 2
- 83
- 154
0
it's doable if you can use $x^4$ as a term in the polynomial. look at $(x^4)^3 + x^4$ and work from there.
mousomer
- 395
- 2
- 7
0
No, you can not. Lets study the function $x^4+x-x^3$ close to zero, that is
$$ x^4+x-x^3 \sim x.$$
Now, you can see that the function always take negative values if we are in the neighbourhood $(-\delta, 0),\quad \delta >0$.
Note: You can make the same analysis for the general function $ax^4+bx-x^3$.
Mhenni Benghorbal
- 47,431