# The question is below?

## If x is real then find the maximum value of $\frac{3 {x}^{2} + 9 x + 17}{3 {x}^{2} + 9 x + 7}$.

Jun 29, 2018

Let

$y = \frac{3 {x}^{2} + 9 x + 17}{3 {x}^{2} + 9 x + 7}$

$\implies y \left(3 {x}^{2} + 9 x + 7\right) = \left(3 {x}^{2} + 9 x + 17\right)$

$\implies 3 \left(y - 1\right) {x}^{2} + 9 \left(y - 1\right) x + \left(7 y - 17\right) = 0$

As $x$ is real we can write

${\left(9 \left(y - 1\right)\right)}^{2} - 4 \cdot 3 \left(y - 1\right) \left(7 y - 17\right) \ge 0$

$\implies 81 {\left(y - 1\right)}^{2} - 4 \cdot 3 \left(y - 1\right) \left(7 y - 17\right) \ge 0$
$\implies 27 {\left(y - 1\right)}^{2} - 4 \left(y - 1\right) \left(7 y - 17\right) \ge 0$

$\implies 27 {y}^{2} - 54 y + 27 - 4 \left(7 {y}^{2} - 24 y + 17\right) \ge 0$

$\implies 27 {y}^{2} - 54 y + 27 - 28 {y}^{2} + 96 y - 68 \ge 0$

$\implies - {y}^{2} + 42 y - 41 \ge 0$

$\implies {y}^{2} - 42 y + 41 \le 0$

$\implies {y}^{2} - 41 y - y + 41 \le 0$

$\implies y \left(y - 41\right) - 1 \left(y - 41\right) \le 0$

$\implies \left(y - 41\right) \left(y - 1\right) \le 0$

This inequality holds for

$1 \le y \le 41$

Hence maximum value of the given expression for real values of x will be $41$

Jun 29, 2018

41

#### Explanation:

$\frac{3 {x}^{2} + 9 x + 17}{3 {x}^{2} + 9 x + 7} = 1 + \frac{10}{3 {x}^{2} + 9 x + 7}$

Hence, to maximize $\frac{3 {x}^{2} + 9 x + 17}{3 {x}^{2} + 9 x + 7}$ we need to minimize $3 {x}^{2} + 9 x + 7$. Now

$3 {x}^{2} + 9 x + 7 = 3 \left({x}^{2} + 3 x\right) + 7$
$q \quad = 3 \left({x}^{2} + 2 \cdot x \cdot \frac{3}{2} + {\left(\frac{3}{2}\right)}^{2}\right) + 7 - 3 {\left(\frac{3}{2}\right)}^{2}$
$q \quad = 2 {\left(x + \frac{3}{2}\right)}^{2} + \frac{1}{4}$

Since ${\left(x + \frac{3}{2}\right)}^{2} \ge 0$ we have

$3 {x}^{2} + 9 x + 7 \ge \frac{1}{4} \implies$
$\frac{10}{3 {x}^{2} + 9 x + 7} \le 40 \implies$

$\frac{3 {x}^{2} + 9 x + 17}{3 {x}^{2} + 9 x + 7} = 1 + \frac{10}{3 {x}^{2} + 9 x + 7} \le 41$

Thus, the maximum value is 41 (attained for $x = - \frac{3}{2}$).