The function f(x) = ax + b x ;a, ,b,x > 0 takes on the least valu… - Vidyadip

MCQ

The function $f(x) = ax + {b \over x};a,\,b,x > 0$ takes on the least value at $x$ equal to

A
$b$
B
$\sqrt a $
C
$\sqrt b $
✓
$\sqrt {b/a} $

✓

Answer

Correct option: D.

$\sqrt {b/a} $

d
(d) $f(x) = ax + \frac{b}{x}$ ==> $f'(x) = a - \frac{b}{{{x^2}}}$

==> $f'(x) = 0 \Rightarrow x = \sqrt {\frac{b}{a}} $

Now $f''(x) = \frac{{2b}}{{{x^3}}}$ ==> At $x = \sqrt {\frac{b}{a}} ,$ $f''(x) = + ve$

$\therefore$ $f(x)$ has the least value at $x = \sqrt {\frac{b}{a}} $.

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