Applications Of Derivatives Question 360
Question: The minimum value of $ x^{2}+\frac{1}{1+x^{2}} $ is at
[UPSEAT 2003]
Options:
A) $ x=0 $
B) $ x=1 $
C) $ x=4 $
D) $ x=3 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ f(x)=x^{2}+\frac{1}{1+x^{2}} $ , $ {f}’(x)=2x-\frac{1}{{{(1+x^{2})}^{2}}},.,2x $
Now $ {f}’(x)=0 $
therefore $ x=0 $ So the function has minimum value at $ x=0 $ .
BETA
