Functions Question 627
Question: If $ f(x)=\frac{x^{2}-1}{x^{2}+1} $ , for every real numbers. then the minimum value of f
[Pb. CET 2001]
Options:
A) Does not exist because f is bounded
B) Is not attained even through f is bounded
C) Is equal to +1
D) Is equal to ?1
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ f(x)=\frac{x^{2}-1}{x^{2}+1}=\frac{x^{2}+1-2}{x^{2}+1}=1-\frac{2}{x^{2}+1} $ $ \because x^{2}+1>1 $ ; \ $ \frac{2}{x^{2}+1}\le 2 $ So $ 1-\frac{2}{x^{2}+1}\ge 1-2 $ ; \ $ -1\le f(x)<1 $ Thus $ f(x) $ has the minimum value equal to ?1.
BETA
