Q.Find the maximum value of the function $\dfrac{x^2 + 14x + 9}{x^2 + 2x + 3}$ over $\mathbb{R}$.

Set $y=f(x)$, differentiate to find critical points, then check the function value at each critical point — the largest is the maximum.

Let $f(x) = \dfrac{x^2+14x+9}{x^2+2x+3}$. Note the denominator $x^2+2x+3=(x+1)^2+2>0$ for all real $x$, so $f$ is defined (and differentiable) everywhere on $\mathbb{R}$.

Step 1 — Differentiate using the quotient rule:

$f'(x) = \dfrac{(2x+14)(x^2+2x+3) - (x^2+14x+9)(2x+2)}{(x^2+2x+3)^2}$

Expanding the numerator:

$(2x+14)(x^2+2x+3) = 2x^3+18x^2+34x+42$

$(x^2+14x+9)(2x+2) = 2x^3+30x^2+46x+18$

Subtracting:

$\text{Numerator} = (2x^3+18x^2+34x+42)-(2x^3+30x^2+46x+18) = -12x^2-12x+24$

Step 2 — Solve $f'(x)=0$:

$-12x^2-12x+24=0 \;\Rightarrow\; x^2+x-2=0 \;\Rightarrow\; (x+2)(x-1)=0 \;\Rightarrow\; x=-2 \text{ or } x=1$

Step 3 — Evaluate $f$ at the critical points:

$f(-2) = \dfrac{4-28+9}{4-4+3} = \dfrac{-15}{3} = -5$

$f(1) = \dfrac{1+14+9}{1+2+3} = \dfrac{24}{6} = 4$

…