Applications Of Derivatives Question 333
Question: The function $ f(x)=ax+\frac{b}{x};a,,b,x>0 $ takes on the least value at x equal to
[AMU 2000]
Options:
A) b
B) $ \sqrt{a} $
C) $ \sqrt{b} $
D) $ \sqrt{b/a} $
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Answer:
Correct Answer: D
Solution:
$ f(x)=ax+\frac{b}{x} $
therefore $ {f}’(x)=a-\frac{b}{x^{2}} $
therefore $ {f}’(x)=0\Rightarrow x=\sqrt{\frac{b}{a}} $ Now $ {f}’’(x)=\frac{2b}{x^{3}} $
therefore At $ x=\sqrt{\frac{b}{a}}, $
$ {f}’’(x)=+ve $ $ f(x) $ has the least value at $ x=\sqrt{\frac{b}{a}} $ .
BETA
