<s>[INST] If the range of $f(x)= \frac{x^2+x+c}{x^2+2x+c}, x\in \mathbb R$ is $\left[\frac 56, \frac 32\right]$ then what is $c$? 
If the range of the function $f(x)= \dfrac{x^2+x+c}{x^2+2x+c}, x\in \mathbb R$ is $\left[\dfrac 56, \dfrac 32\right]$ then $c$ is equal to? 

Attempt: 
$y=  \dfrac{x^2+x+c}{x^2+2x+c}$
For real values of $x$, $\Delta \ge 0$
$\implies 4y^2(1-4c) +1-4y(1-2c) - 4c \ge 0$
What do I do next? I am really unable to understand the concept to be followed after this. Could someone explain that? 
The answer is: 

 $c= 4$

 [/INST] $$f'(x)=\frac{x^2-c}{(x^2+2x+c)^2}=0 \implies x=\pm \sqrt c$$
$$f(\pm \sqrt c)\in \{3/2, 5/6\}$$
$$f(\sqrt{c})=1-\frac{1}{2(\sqrt{c}+1)} = 5/6\implies c=4$$
$$f(-\sqrt{c})=1+\frac{1}{2(\sqrt{c}-1)} =3/2      \implies c=4    $$
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