What is the range of the function $f(x) = \frac{7x^{2}-4x+4}{x^{2}+1}$?

Question

What is the range of $f(x) = \frac{7x^{2}-4x+4}{x^{2}+1}$? I can't find a method to solve this question. I know the answer, but I don't know how to solve it. Any help would be appreciated.

Answer 1

Let $y= \frac{7x^{2}-4x+4}{x^{2}+1}$ $\Rightarrow (y-7)x^2+4x+(y-4)=0$ As $x$ must be real, the discriminant is $≥0$ $\Rightarrow 16-4(y-7)(y-4)≥0$ $\Rightarrow y^2-11y+24≤0$

Can you finish?

Answer 2

$f(x)=7-\frac{4x+3}{x^2+1}=7-g(x)$

We now have to find the range of $g(x)=\frac{4x+3}{x^2+1}$. We can see that the domain of this is $(-\infty,\infty)$ and that there are no vertical asymptotes and a horizontal asymptote of $y=0$

We can do some more inspection of this function and see that for $x>-\frac{3}{4}$, $g(x)>0$ and for $x<-\frac{3}{4}$, $g(x)<0$. So this means that $g(x)$ looks something like the derivative of a negative normal distribution. Specifically this means that there will be a local maxima at $x>-\frac{3}{4}$ and a local minima at $x<-\frac{3}{4}$.

$g'(x)=\frac{4(x^2+1)-8x^2-6x}{(x^2+1)^2}$. Solving for the location of extrema, we get $2x^2+3x-2=0$, which has solutions $x=\{-2,0.5\}$. We already determined which one was the minima and the maxima, so the range of $g(x)$ is $[g(-2),g(.5)]=[-1,4]$

So the range of $f(x)$ is $[3,8]$.