<s>[INST] Prove that $f(x)=\frac{ax^2+x-2}{a+x-2x^2}$ has the range $ℝ$ when $x\inℝ$ if $a \in [1,3]$ Prove that $f(x)=\frac{ax^2+x-2}{a+x-2x^2}$ has the range $ℝ$ when $x\inℝ$ if $a \in [1,3]$
My working:
Let  $f(x)=\frac{ax^2+x-2}{a+x-2x^2}=y$
therefore,
$$(a+2y)x^2+(1-y)x-(2+ay)=0$$
Now, for $x$ to have real values, the discriminant of the above equation must not be less than zero.
So,
$$(1+8a)y^2+(14+4a^2)y+(1+8a)\ge 0$$
Now, somehow from this, we have to land on the condition for $a$ which I am unable to do.
 [/INST] The minimum of the polynomial $ax^2+bx+c$ , where $a$ is positive, is $\frac{4ac - b^2}{4a}$. To see this, note that $ax^2 + bx + c = \frac{(4a^2x^2 + 4abx + b^2)}{4a} -\frac{b^2}{4a} + c =  \frac 1{4a} (2ax + b)^2 + \frac{4ac - b^2}{4a}$, so the smallest value is when the square is zero, then the value is $\frac{4ac - b^2}{4a}$.
In other words, the polynomial is positive everywhere if and only if $4ac > b^2$.
Note that if $1 \leq a \leq 3$, then $(1+8a) > 0$, so you can apply the given criterion to conclude that it is enough to check that $4 \times (1+8a) \times (1+8a) > (14 + 4a^2)^2$ whenever $1 \leq a \leq 3$.
Can you check this?
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