Example 3.34 (Page 97 in my version) in their book defines $$\mathrm{PV}(x,r):=\sum_{i=1}^n (1+r)^{-i}x_i$$ for every $r\geq 0$ and $x \in \Bbb R^n$ such that $x_0 < 0$ and $\sum_{i=1}^n x_i>0$. Then, they define $$\mathrm{IRR}(x)=\inf\{r\geq 0\mid \mathrm{PV}(x,r)=0 \}$$ and claim, for every $x$ in the above domain, $$\mathrm{IRR}(x) \geq R \Leftrightarrow \forall r\in [0,R): \mathrm{PV}(x,r)>0.$$
I see that $$\mathrm{IRR}(x)\geq R \Leftrightarrow \inf\{r\geq 0\mid \mathrm{PV}(x,r)=0 \} \geq R\\ \Leftrightarrow \forall r\geq 0: (\mathrm{PV}(x,r)=0 \Rightarrow r\geq R)\\ \Leftrightarrow \forall r\in [0,R) \Rightarrow \mathrm{PV}(x,r)\neq 0.$$
As stated there, $PV(x,0)>0$ and $PV(x,r)<0$ where $r\to \infty$. Still, I can't see why we cannot have more than one root in between without additional argument.
We have $$\forall r\in [0,R): \mathrm{PV}(x,r)\neq 0 \Rightarrow \forall r\in [0,R): \mathrm{PV}(x,r)>0$$ by the Intermediate value theorem, and the fact $\mathrm{PV}(x,0)>0$. The other direction trivially holds.