Description for Correct answer:
Let \( \Large y=\frac{x^{2}-3x+4}{x^{2}+3x+4} \)

=> \( \Large \left(y-1\right)x^{2}+3 \left(y+1\right)x+4 \left(y-1\right)=0 \)

For x is real, \( \Large D \ge 0 \)

=> \( \Large 9 \left(y+1\right)^{2}-16 \left(y-1\right)^{2}\ge 0 \)

=> \( \Large -7y^{2}+50y-7\ge 0 \)

=> \( \Large 7y^{2}-50y+7 \le 0 \)

=> \( \Large \left(y-7\right) \left(7y-1\right) \le 0 \)

=> \( \Large y \le 7\ and\ y \ge \frac{1}{7} => \frac{1}{7} \le y \le 7 \)

Hence maximum value 7 and minimum is \( \Large \frac{1}{7} \).