muroscamsey

Answered

2022-08-12

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

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

$f(a)>u>f(b)$

$f(0)>0>f(1)$

$p>0>p-4$

This is only true for $p:(0,4)$ ie $0<p<4$

Now I have proved that there is atleast one root between $[0,1]$, 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?

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

$f(a)>u>f(b)$

$f(0)>0>f(1)$

$p>0>p-4$

This is only true for $p:(0,4)$ ie $0<p<4$

Now I have proved that there is atleast one root between $[0,1]$, 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?

Answer & Explanation

Hamza Conrad

Expert

2022-08-13Added 20 answers

Yes, Rolle's theorem is perhaps the best tool for this. If $f$ had two zeros, then the derivative

$f(x)=5{x}^{4}-5=5({x}^{4}-1)$

would vanish at a point in the interior of $[0,1]$ - however, this isn't possible since ${x}^{4}-1$ is negative for all such $x$.

$f(x)=5{x}^{4}-5=5({x}^{4}-1)$

would vanish at a point in the interior of $[0,1]$ - however, this isn't possible since ${x}^{4}-1$ is negative for all such $x$.

moiraudjpdn

Expert

2022-08-14Added 2 answers

First you can see that ${f}^{\prime}(x)=5({x}^{4}-1)$ 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!

Most Popular Questions