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

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

(Can we solve it Using $\bf{A.M\geq G.M}$) Inequality.

$\bf{My\; Try::}$ Let $\displaystyle y = f(x) = \frac{x^2+14x+9}{x^2+2x+3} = \frac{(x+1)^2+12(x+1)-4}{(x+1)^2+1}$

Now Let $(x+1) = t\;,$ where $t\in \mathbb{R}$

So $\displaystyle y=\frac{t^2+12t-9}{t^2+1} = 1+\frac{12t-10}{t^2+1}$

Now I did not understand How can I solve after that

Help me

Thanks

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 $$+4$$...

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 $\ge0$