If $f:[\alpha,\beta]\to\mathbb{R}$ has positive derivative and $f(a)\cdot f(b)<0$, then is there a unique root?

Question

Reading Zorich, mathematical analysis II, pag 38 (introduction on Newton's method) I found this sentence:

My question is: is the convexity really required to say only that there is a unique point $a\in[\alpha,\beta]$ such that $f (a) = 0$?

Answer 1

As pointed out in the comments, positive derivative implies strictly increasing. Therefore $f(a) = 0$ for at most one $a \in [\alpha,\beta]$. We have $f(a) = 0$ for exactly one $a \in [\alpha,\beta]$ if $f(\alpha) < 0$ and $f(\beta) > 0$. The condition that $f$ assumes values of opposite sign at the endpoints of the interval is formally correct, but misleading. Since $f$ is strictly increasing, the case $f(\alpha) > 0 , f(\beta) < 0$ is impossible.