<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] Note:
$$y=  \dfrac{x^2+x+c}{x^2+2x+c} \Rightarrow \\ (y-1)x^2+(2y-1)x+c(y-1)=0.$$
For $x$ to be real:
$$\Delta=(2y-1)^2-4c(y-1)^2=4(1-c)y^2+2(4c-2)y+1-4c\ge0.$$
The range $y\in\left[\frac65,\frac32\right]$ corresponds to the solution of the above inequality with $y_1=\frac56$, $y_2=\frac32$ and $y_1y_2=\frac54$: 
$$y_1y_2=\frac{1-4c}{4(1-c)}=\frac54 \Rightarrow c=4.$$
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