Given-f-x-1-2x-x-2-1-4x-2-3x-3-2x-find-f-2-

Question Number 105822 by bemath last updated on 01/Aug/20

G i v e n f (x + \frac{1}{2 x}) = x^{2} + \frac{1}{4 x^{2}} + 3 x + \frac{3}{2 x}

f i n d f (2)

Answered by bobhans last updated on 01/Aug/20

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

\Leftrightarrow f (x + \frac{1}{2 x}) = {(x + \frac{1}{2 x})}^{2} + 3 (x + \frac{1}{2 x}) - 1

\Leftrightarrow f (X) = X^{2} + 3 X - 1

f (2) = 4 + 6 - 1 = 9 ▴

Answered by mathmax by abdo last updated on 01/Aug/20

let x + \frac{1}{2 x} = t \Rightarrow {2 x}^{2} + 1 = 2 tx \Rightarrow {2 x}^{2} - 2 tx + 1 = 0 \Rightarrow

Δ^{^{'}} = t^{2} - 2 \Rightarrow x_{1} = \frac{t + \sqrt{t^{2} - 2}}{2} and x_{2} = \frac{t - \sqrt{t^{2} - 2}}{2}

x = x_{1} \Rightarrow f (t) = {(\frac{t + \sqrt{t^{2} - 2}}{2})}^{2} + \frac{1}{4 {(\frac{t + \sqrt{t^{2} - 2}}{2})}^{2}} + 3 \times \frac{t + \sqrt{t^{2} - 2}}{2}

+ \frac{3}{2 (\frac{t + \sqrt{t^{2} - 2}}{2})} \Rightarrow

f (2) = {(\frac{2 + \sqrt{2}}{2})}^{2} + \frac{1}{{(2 + \sqrt{2})}^{2}} + 3 . \frac{2 + \sqrt{2}}{2} + \frac{3}{2 + \sqrt{2}}

x = x_{2} \Rightarrow f (t) = {(\frac{t - \sqrt{t^{2} - 2}}{2})}^{2} + \frac{1}{4 {(\frac{t - \sqrt{t^{2} - 2}}{2})}^{2}} + 3 \times \frac{t - \sqrt{t^{2} - 2}}{2}

+ \frac{3}{2 (\frac{t - \sqrt{t^{2} - 2}}{2})} \Rightarrow

f (2) = {(\frac{2 - \sqrt{2}}{2})}^{2} + \frac{1}{4 {(\frac{2 - \sqrt{2}}{2})}^{2}} + \frac{3}{2} (2 - \sqrt{2}) + \frac{3}{2 - \sqrt{2}}