muroscamsey

2022-08-12

Let $f\left(x\right)={x}^{5}-5x+p$. Show that $f$ can have at most one root in $\left[0,1\right]$,regardless of the value of p.

This seems to be an IVT problem, so I will go forth with that:

$f\left(a\right)>u>f\left(b\right)$
$f\left(0\right)>0>f\left(1\right)$
$p>0>p-4$
This is only true for $p:\left(0,4\right)$ ie $0
Now I have proved that there is atleast one root between $\left[0,1\right]$, now I need to prove that there isn't a second.

Would Rolles theorem be what I need here, or is there a simple deduction I can make from above to finish the problem?

Expert

Yes, Rolle's theorem is perhaps the best tool for this. If $f$ had two zeros, then the derivative
$f\left(x\right)=5{x}^{4}-5=5\left({x}^{4}-1\right)$
would vanish at a point in the interior of $\left[0,1\right]$ - however, this isn't possible since ${x}^{4}-1$ is negative for all such $x$.

moiraudjpdn

Expert

First you can see that ${f}^{\prime }\left(x\right)=5\left({x}^{4}-1\right)$ is negative for the given interval. Therefore, the given function is a decreasing function. So, depending upon the value of the function at 0 and 1 it will have one or no roots. Hope this helps!