<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] Write $f(x)=1-\frac x{x^2+2x+c}$  As $x \to \pm \infty$ this goes to $1$.  If the denominator has a real root it will go off to $\pm \infty$ so $c \gt 1$.  The denominator is then always positive, so the cases where $f(x) \gt 1$ are where $x \lt 0$ and the cases where $f(x) \lt 1$ are where $x \gt 0$.  Take the derivative, set to zero, and find the $x$ value of the local maximum as a function of $c$.  Plug that $x$ value into $f(x)$ and set it equal to $\frac 32$.  You will get an equation for $c$.  You can then check that the minimum has $f(x)=\frac 56$
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