# f(x) = x^5−15x^4−10x^2+20 Prove that the equation f(x)=0 has a real solution with x>=1.

$f\left(x$) = ${x}^{5}-15{x}^{4}-10{x}^{2}+20$
Prove that the equation $f\left(x\right)=0$ has a real solution with $x\ge 1$.
Idea: Started off by coming up with a boundary $\left[1,x\right]$ for some real number $x$. $f\left(1\right)=-4$ so is less than zero, then attempted to find an x that gave me a positive result, very quickly came to the conclusion that any number above 1 when inserted into a function gave me a negative number. Pretty sure I could be missing something really simple. Any help or hints would be appreciated.
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yermarvg
$f\left(20\right)$ is positive. So, check again.
Also, for future purposes, to solve such questions, you can easily check the behaviour of a polynomial at large $x$ by just considering the sign of the highest power variable. If it is positive, then $f\left(x\right)\to -\mathrm{\infty }$, and if it is negative, then $f\left(x\right)\to -\mathrm{\infty }$.
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Raynor2i
Hint: $f\left(1\right)<0$ and $\underset{x\to \mathrm{\infty }}{lim}=\mathrm{\infty }.$