Please help!! PROVE THAT f(x) = 2(1-x)(1+x) 1+x) ≤ 64/27 . and reaches maximum value for x ∈[0,1] off x=1/3
Can anyone tell me what the steps are, I just keep getting 4=4 whenever I try to solve it!
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@arberavdullahu I'm getting stuck, it comes to 4=4 can you please help? It'll be really appreciated! – user135643 Nov 12 '16 at 10:25
1 Answers
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If you use AM-GM for $2-2x,1+x,1+x$(we can use it because all of them are non negative) you get that $$\frac{2-2x+1+x+1+x}{3}\geq \sqrt[3]{(2-2x)(1+x)(1+x)}$$ or
$$\frac{4}{3}\geq \sqrt[3]{(2-2x)(1+x)(1+x)}$$
getting ride of cube we get that
$$\frac{64}{27}=(\frac{4}{3})^3\geq (2-2x)(1+x)(1+x)$$
So $$2(1-1x)(1+x)(1+x)\leq \frac{64}{27}$$
Also it reaches maximum in case of equality i.e. $2-2x=1+x$ so $x=\frac{1}{3}$
Let me know if you don't get any part of it.
arberavdullahu
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Wow thank you, just one question how do we know that they are non negative? – user135643 Nov 12 '16 at 10:48
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You are welcome. Yeah basically you look it's minimum for example for 2-2x minimum is when x is maximum so x=1 then minimum of 2-2x = 0, similiar for 1+x – arberavdullahu Nov 12 '16 at 10:52
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how do I prove that it reaches maximum value for x belonging to [0,1] off x=1/3, I mean how to write it exactly? Sorry for asking too much – user135643 Nov 12 '16 at 10:59
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Hey can you please help me with a question I just asked , whenever you are free – user135643 Nov 13 '16 at 12:44