Find the set of values of a for which $(a-1)x^2 - (a+1)x +a-1\ge 0$ is true for all $x\ge 2$
There are two constraints for the condition to be satisfied:
- $a-1>0 \implies a>1$
- $\Delta\le0$ which yields $a\in(-\infty,1/3]\cup[3,\infty)$
But I am unable to understand how to use the condition $x\ge2$. Can someone guide me on that?
PS: Answer is: $[7/3,\infty)$
The constraint $x \ge 2$ is used to figure out a constraint on $a$. In particular, for $x=2$, $$(a-1)x^2-(a+1)x+(a-1) \ge 0 \implies a \ge 7/3.$$
Further, the minimum of $\alpha x^2+\beta x+\gamma$ occurs at $x = -\beta/(2\alpha)$ when $\alpha > 0$. In your case, $$-\frac{\beta}{2\alpha}=\frac{a+1}{2(a-1)} \le 2 \implies a\ge 5/3.$$ Consequently, the minimum of $(a-1)x^2-(a+1)x+(a-1)$ occurs before $x=2$ for $a \ge 7/3$, which implies the function increases monotonically for $x\ge 2$. As such, $\color{blue}{a \ge 7/3}$ implies $(a-1)x^2-(a+1)x+(a-1) \ge 0$ for $x \ge 2$.