Show $f_t(x)=x(1+t(1-x))<1$

Question

How can I show that $f_t(x)=x(1+t(1-x))<1$ for $t\in (0,1]$ arbitrary and every $x\in (0,1)$? Thanks for a hint, I only got $\leq$.

Answer 1

As far as I can see, you want to solve it without recourse to differentiation, which will identify when a given differentiable function is increasing. So, here is a simpler solution: Clearly, \begin{equation*} f_{t}(x)=x(1+t(1-x))<1 \end{equation*} holds if and only if \begin{equation*} 0<1-x(1+t(1-x))=(1-x)(1-tx)\text{,} \end{equation*} where it is straightforward to see that $0<1-x$ and $0<1-tx$ whenever $x\in (0,1)$ and $t\in (0,1]$.

Hope this helps.