Range of the function $f(x) = \frac{x^2+14x+9}{x^2+2x+3}\;,$ where $x\in \mathbb{R}$

Solution 1:

Hint:

try to find a line $y=k$ such that

$$ \frac{x^2+14x+9}{x^2+2x+3} - k = 0 $$

has only one solution, that $k$ is a max (above the max there is no solution, below there are two solutions) or a min (below the min there is no solution, abore there are two solutions).

Thus

$$ \frac{[1-k]x^2+2[7-k]x+[9-3k]}{x^2+2x+3} = 0 $$

Set discriminant to $0$ and find $k$...

So we get$$[7-k]^2 - [1-k][0-3k] = 0,$$ whence $$-2k^2-2k+40=0,$$ so $$k^2+k-20=0,$$ thus $$[k+5][k-4]=0,$$ so the min is $-5$ and the max is $+4$...

Solution 2:

HINT:

Let $\displaystyle y=\frac{x^2+14x+9}{x^2+2x+3}$

Rearrange to form a Quadratic Equation in $x$

As $x$ is real, the discriminant must be $\ge0$