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:
$c= 4$
On
Hint. Note that $$f'(x)=\frac{x^2-c}{(x^2+2x+c)^2}\quad\mbox{and}\quad \lim_{x\to \pm \infty}f(x)=1.$$ Moreover $f$ is bounded over the real line if and only if $x^2+2x+c=(x+1)^2+c-1>0$ that is $c>1$. In the bounded case, in order to determine the range of the continuous function $f$, consider the values of $f$ at $\pm \sqrt{c}$ (where the derivative $f'$ is zero): $$f(\sqrt{c})=1-\frac{1}{2(\sqrt{c}+1)}<1<f(-\sqrt{c})=1+\frac{1}{2(\sqrt{c}-1)}.$$
On
$$f'(x)=\frac{x^2-c}{(x^2+2x+c)^2}=0 \implies x=\pm \sqrt c$$
$$f(\pm \sqrt c)\in \{3/2, 5/6\}$$
$$f(\sqrt{c})=1-\frac{1}{2(\sqrt{c}+1)} = 5/6\implies c=4$$
$$f(-\sqrt{c})=1+\frac{1}{2(\sqrt{c}-1)} =3/2 \implies c=4 $$
On
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 $y_1y_2=\frac54$: $$y_1y_2=\frac{1-4c}{4(1-c)}=\frac54 \Rightarrow c=4.$$
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 $f(x)=\frac 56$