Applications Of Derivatives Question 343
Question: If $ f(x)=x+\frac{1}{x}, $ x > 0, then its greatest value is
[RPET 2002]
Options:
A) - 2
B) 0
C) 3
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ f(x)=x+\frac{1}{x} $
therefore $ {f}’(x)=1-\frac{1}{x^{2}} $
Put $ {f}’(x)=0 $ , $ x=-1,,1 $
Since x > 0, so no maximum value can be found.
BETA
