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I'm working on a optimization problem and need to show that \begin{equation} \frac{1}{2}x^4 - x^3 -x + 100 = 0 \end{equation} has no real solution in order to prove certain properties about the function I am working on. I don't need to know the zeros just show that they don't exist. I'm doing this because I would like to show that \begin{equation} \frac{1}{2}x^4 - x^3 > x - 100 \end{equation} for all $x$. What would be the best way to go about this?

Thanks in advance.

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see… there I use Sturm's theorem to show this. This method is straightforward and is implemented in some CAS (e.g. Maxima) to find the numbers of real roots in an interval – miracle173 Feb 3 '14 at 0:02
up vote 5 down vote accepted

We can do it by looking at the expression for different ranges of $x$. It is clearly positive if $x\le 0$.

If $0 < x \le 4$, then $x^3+x\lt 100$, so our expression is positive even without the aid of $\frac{1}{2}x^4$. If $x\gt 4$, then $\frac{1}{2}x^4\gt 2x^3$, so $\frac{1}{2}x^4-x^3-x\gt 0$.

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Thank you. Much appreciated. – user245465 Feb 2 '14 at 18:27
You are welcome. – André Nicolas Feb 2 '14 at 18:31

This is a quartic, its discriminant tells all you need.

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For negative $x$, all summands are $\ge 0$ so $f(x)\ge 100$. For $0\le x \le 4$, $f(x)\ge 100-x^3-x\ge 100-64-4>0$. Note that $$ f(x+1)=\frac12x^4+x^3-2x+\frac{197}2\ge (x^2-2)x+\frac{197}2>0$$ for $x>\sqrt 2$, so $f(x)>0$ for $x>\sqrt 2-1$. This covers all of $\mathbb R$.

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Find the derivative ov the left side of your first equation and use it's zeros to find the minima of that function. They will, hopefully, all be positive :)

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Sketch $\frac{1}{2}x^4 - x^3 -x + 100$. There are two changes in concavity between $x=-2$ and $x=3$, so the graph can't "change direction" outside those values, and the minimum must be between them. Direct calculation shows that within that interval, the function exceeds 50 at each integer-valued $x$.

For there to be a zero within $0.5$ units of any of these integer points, the function would have to have an average rate of change of more than $50$ on an interval of length less than $0.5$, and by the mean value theorem, the absolute value of the derivative would have to exceed $\frac{50}{0.5}=100$ somewhere. It doesn't. The absolute value of the derivative in that interval is easily bounded above by $2|3|^3+3|3|^2+1=82$ (the sum of the absolute values of the terms of the derivative at the largest $|x|$-value.

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