Applications Of Derivatives Question 269
Question: If $ f(x)=\frac{x^{2}-1}{x^{2}+1} $ , for every real number x, then the minimum value of f
[IIT 1998]
Options:
A) Does not exist because f is unbounded
B) Is not attained even though f is bounded
C) Is equal to 1
D) Is equal to -1
Show Answer
Answer:
Correct Answer: D
Solution:
$ f(x)=\frac{x^{2}-1}{x^{2}+1}=\frac{x^{2}+1-2}{x^{2}+1}=1-\frac{2}{x^{2}+1} $
$ \therefore f(x)<1\forall x $ and $ \ge -1 $ as $ \frac{2}{x^{2}+1}\le 2 $
$ \therefore -1\le f(x)<1 $
Hence $ f(x) $ has minimum value -1 and also there is no maximum value. Aliter : $ {f}’(x)=\frac{(x^{2}+1)2x-(x^{2}-1)2x}{{{(x^{2}+1)}^{2}}}=\frac{4x}{{{(x^{2}+1)}^{2}}} $
$ {f}’(x)=0\Rightarrow x=0 $
$ {f}’’,(x)=\frac{{{(x^{2}+1)}^{2}}4-4x.2(x^{2}+1)2x}{{{(x^{2}+1)}^{4}}} $
$ =\frac{(x^{2}+1)4-16x(x)}{{{(x^{2}+1)}^{3}}}=\frac{-12x^{2}+4}{{{(x^{2}+1)}^{3}}} $
$ \therefore {f}’’(0)>0 $
$ \therefore $ There is only one critical point having minima.
Hence $ f(x) $ has least value at $ x=0 $ . $ {f _{\min }}=f(0)=\frac{-1}{1}=-1 $ .
BETA
