Question

find the max value of the expression: $f\left(x\right)=\frac{x^2-2}{x^2-3}$

• A. $0$
• B. $\infty$
• C. $\frac{2}{3}$
• D. $1$

Answer 1

Answer: (B)

$\infty$

$f\left(x\right)=\frac{x^2-2}{x^2-3}=\frac{x^2-3+1}{x^2-3}=1+\frac{1}{x^2-3}$
Thus maximum value of $f\left(x\right)$ is $\infty$ at $x=\pm \sqrt{3}$