Find $m$ such that $ f(x)<0,x∈(0,1)$

Question

Let $f ( x ) = x ^ { 2} - ( m - 1) x + 3m - 4,m \in R$. Find m such that $f(x)<0, x∈(0,1)$. Knowing that the interval between the roots of $f(x)=0$ must have negative values through f, I have tried to imply that $x _ { 1} \leq 0,x _ { 2} \geq 1$. This led me to $m≤4/3,m≤13/10$, however, which is the wrong answer.

Answer 1

Hint For any strictly convex function $g$, if $x_0 < x < x_1$, then $$g(x) < \max\{g(x_0), g(x_1)\} .$$ So, it is enough to determine the behavior of $f$ at the endpoints $0, 1$ of the given interval.