<s>[INST] Find $a$ if $(a+3)x^2-4x+2<0$ $\forall x\in [-2,1]$ Find $a$ if $(a+3)x^2-4x+2<0 \qquad \forall x\in [-2,1]$
My Attempt:
$a+3=0$ $\implies -4x+2<0$
$$2x-1>0$$
$$x>\dfrac {1}{2}$$
$\implies a \neq -3$.
 [/INST] HINT
Let $f(x)= (a+3)x^2-4x+2$
Firstly, we discuss for $a+3$ is positive, negative, or zero.
Then for each case, we check for $x=-\frac{-4}{2(a+3)}=\frac{2}{a+3}$, is it on the left of $[-2,1]$, on the right, or in between?
For example, if $a+3 < 0$:
If $\frac{2}{a+3} \le -2$: it is decreasing, and we just want $f(-2) <0$
If $-2 < \frac{2}{a+3} <1$: we need its local maximal to be negative, $f(\frac{2}{a+3}) < 0$
If $\frac{2}{a+3} \ge 1$: it is increasing, $f(1) <0$
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