Question

Show that $\frac{1}{7}\le \frac{x^2-3x+4}{x^2+3x+4}\le 7$.

Answer 1

$y\left(say\right)=$$\frac{x^2-3x+4}{x^2+3x+4}$

$(y-1)x^2+(3y+3)x+4y-4=0$

For real solutions,$D\ge 0$

$\Rightarrow D\equiv \left[3\right(y+1){]}^2+16(y-1{)}^2\ge 0$

$\Rightarrow 9y^2+18y+9+16y^2-32y+16\ge 0$

$\Rightarrow 25y^2-14y+25\ge 0$

$D=196-2500<0$

$\Rightarrow$ equation$>0\forall y\in R$ as $a>0$

Now, $y\in R,\Rightarrow y$ also $\in$ each and every subset of it.

$\Rightarrow \frac{1}{7}\le y\le 7$