SATHEE: Applications Of Derivatives Question 343

Applications Of Derivatives Question 343

If $ f(x)=x+\frac{1}{x}, $ x > 0, then its minimum value is

[RPET 2002]

Options:

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, there is no maximum value.

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Applications Of Derivatives Question 343

If $ f(x)=x+\frac{1}{x}, $ x > 0, then its minimum value is

Options:

Answer:

Solution:

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