Intermediate value theorem problem(proving only one root)

Question

This problem is likely very trivial.

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:

$f(a) \gt u \gt f(b)$

$f(0) \gt 0 \gt f(1)$

$p \gt 0 \gt p-4$

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

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 1

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

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

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 $x$.

Answer 2

First you can see that $f'(x) = 5(x^4-1)$ 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!