f(x) = (1/(1+2(√(2x))+2x))+(x/4) prove that f(x) ≥ (3/8) .


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Question Number 79588 by loveineq. last updated on 26/Jan/20

$f (x) = \frac{1}{1 + 2 \sqrt{2 x} + 2 x} + \frac{x}{4}$ $prove that f (x) ⩾ \frac{3}{8} .$
Commented by MJS last updated on 26/Jan/20

$2 \sqrt{2 x} or 2 \sqrt{2} x ?$
Commented by loveineq. last updated on 26/Jan/20

$is 2 \sqrt{2 x} not 2 \sqrt{2} x$
Commented by john santu last updated on 26/Jan/20
$let (√(2x)) = t ⇒2x = t^2 f(x) = (1/(1+2t+t^2 )) + (t^2 /8) f(x) = (1/((t+1)^2 )) +(t^2 /8) = (t+1)^(−2) +(t^2 /8) f ′(x) = {−2(t+1)^(−3) +(t/4)}×(1/t)=0 ((−2)/((t+1)^3 ))+(t/4)=0 ⇒t(t+1)^3 =8 t = 1 ⇒x = (1/2) ⇒f_(min) = (1/(1+2(√(2.(1/2)))+2.(1/2)))+(1/8) = (1/(1+2+1))+(1/8)=(3/8)$
$let \sqrt{2 x} = t \Rightarrow 2 x = t^{2}$ $f (x) = \frac{1}{1 + 2 t + t^{2}} + \frac{t^{2}}{8}$ $f (x) = \frac{1}{{(t + 1)}^{2}} + \frac{t^{2}}{8} = {(t + 1)}^{- 2} + \frac{t^{2}}{8}$ $f^{'} (x) = {- 2 {(t + 1)}^{- 3} + \frac{t}{4}} \times \frac{1}{t} = 0$ $\frac{- 2}{{(t + 1)}^{3}} + \frac{t}{4} = 0 \Rightarrow t {(t + 1)}^{3} = 8$ $t = 1 \Rightarrow x = \frac{1}{2} \Rightarrow f_{\min} =$ $\frac{1}{1 + 2 \sqrt{2 . \frac{1}{2}} + 2 . \frac{1}{2}} + \frac{1}{8}$ $= \frac{1}{1 + 2 + 1} + \frac{1}{8} = \frac{3}{8}$
	Answered by MJS last updated on 26/Jan/20

	$\frac{1}{2 x + 2 \sqrt{2} x^{\frac{1}{2}} + 1} + \frac{x}{4} < \frac{3}{8}$ $\Rightarrow$ $x^{2} + \sqrt{2} x^{\frac{3}{2}} - x - \frac{3 \sqrt{2}}{2} x^{\frac{1}{2}} + \frac{5}{4} < 0$ ${(\sqrt{x})}^{4} + \sqrt{2} {(\sqrt{x})}^{3} - {(\sqrt{x})}^{2} - \frac{3 \sqrt{2}}{2} \sqrt{x} + \frac{5}{4} < 0$ ${(\sqrt{x} - \frac{\sqrt{2}}{2})}^{2} ({(\sqrt{x})}^{2} + 2 \sqrt{2} \sqrt{x} + \frac{5}{2}) < 0$ $wrong \Rightarrow \frac{1}{2 x + 2 \sqrt{2} x^{\frac{1}{2}} + 1} + \frac{x}{4} ⩾ \frac{3}{8}$
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