Given-a-function-f-x-x-2-1-x-2-4x-4-x-where-x-gt-0-find-the-minimum-value-of-f-x-

Question Number 114181 by bemath last updated on 17/Sep/20

G i v e n a f u n c t i o n

f (x) = x^{2} + \frac{1}{x^{2}} + 4 x + \frac{4}{x}; w h e r e x > 0 .

f i n d t h e m i n i m u m v a l u e o f f (x)

Answered by bobhans last updated on 17/Sep/20

r e c a l l p + \frac{1}{p} ⩾ 2; p > 0

n o w f (x) = (x^{2} + \frac{1}{x^{2}}) + 4 (x + \frac{1}{x})

\Rightarrow f (x) = {(x + \frac{1}{x})}^{2} - 2 + 4 (x + \frac{1}{x})

f (x) ∣_{m i n i m u m} = 2^{2} - 2 + 4.2 = 10 .

Answered by mathmax by abdo last updated on 17/Sep/20

x^{2} + \frac{1}{x^{2}} ⩾ 2 and x + \frac{1}{x} ⩾ 2 \Rightarrow 4 (x + \frac{1}{x}) ⩾ 8 \Rightarrow x^{2} + \frac{1}{x^{2}} + 4 (x + \frac{1}{x}) ⩾ 10 \Rightarrow

\min_{R^{+}} f (x) = 10

Answered by 1549442205PVT last updated on 18/Sep/20

f (x) = x^{2} + \frac{1}{x^{2}} + 4 (x + \frac{1}{x})

= {(x - \frac{1}{x})}^{2} + 4 {(\sqrt{x} - \frac{1}{\sqrt{x}})}^{2} + 10 ⩾ 10

The equality ocurrs if and only if

x = 1 . Hence, f {(x)}_{\min} = 10 when x = 2