Maximum value of function $f(x)=frac{x^4-x^2}{x^6+2x^3-1}$ when $x >1$
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What is the maximum value of the $$f(x)=frac{x^4-x^2}{x^6+2x^3-1}$$ where $x > 1$ .
My try
Unable to solve further.
real-analysis functions optimization maxima-minima a.m.-g.m.-inequality
asked Jan 16 at 11:40
catttcattt
242
$endgroup$
add a comment |
$begingroup$
What is the maximum value of the $$f(x)=frac{x^4-x^2}{x^6+2x^3-1}$$ where $x > 1$ .
My try
Unable to solve further.
real-analysis functions optimization maxima-minima a.m.-g.m.-inequality
asked Jan 16 at 11:40
catttcattt
242
$endgroup$
$begingroup$
What are $a$ and $b$ on the second page?
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– Matti P.
Jan 16 at 11:43
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Hint: the derivative can be written as $$ f'(x) = frac{-2x(x^2 +1)(x^2-x-1)(x^4 + x^3 - x^2 - x -1)}{(x^6 + 2x^3 -1)^2} $$ Source: wolframalpha.com/input/…
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– Matti P.
Jan 16 at 11:46
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$f(x)$ would be maximum when $1/f(x)$ would be minimum.
$endgroup$
– Paras Khosla
Jan 16 at 14:11
add a comment |
$begingroup$
What is the maximum value of the $$f(x)=frac{x^4-x^2}{x^6+2x^3-1}$$ where $x > 1$ .
My try
Unable to solve further.
real-analysis functions optimization maxima-minima a.m.-g.m.-inequality
asked Jan 16 at 11:40
catttcattt
242
$endgroup$
What is the maximum value of the $$f(x)=frac{x^4-x^2}{x^6+2x^3-1}$$ where $x > 1$ .
My try
Unable to solve further.
real-analysis functions optimization maxima-minima a.m.-g.m.-inequality
real-analysis functions optimization maxima-minima a.m.-g.m.-inequality
asked Jan 16 at 11:40
catttcattt
242
asked Jan 16 at 11:40
catttcattt
242
edited Jan 16 at 15:48
cattt
asked Jan 16 at 11:40
catttcattt
242
asked Jan 16 at 11:40
catttcattt
242
asked Jan 16 at 11:40
catttcattt
242
242
$begingroup$
What are $a$ and $b$ on the second page?
$endgroup$
– Matti P.
Jan 16 at 11:43
$begingroup$
Hint: the derivative can be written as $$ f'(x) = frac{-2x(x^2 +1)(x^2-x-1)(x^4 + x^3 - x^2 - x -1)}{(x^6 + 2x^3 -1)^2} $$ Source: wolframalpha.com/input/…
$endgroup$
– Matti P.
Jan 16 at 11:46
$begingroup$
$f(x)$ would be maximum when $1/f(x)$ would be minimum.
$endgroup$
– Paras Khosla
Jan 16 at 14:11
add a comment |
$begingroup$
What are $a$ and $b$ on the second page?
$endgroup$
– Matti P.
Jan 16 at 11:43
$begingroup$
Hint: the derivative can be written as $$ f'(x) = frac{-2x(x^2 +1)(x^2-x-1)(x^4 + x^3 - x^2 - x -1)}{(x^6 + 2x^3 -1)^2} $$ Source: wolframalpha.com/input/…
$endgroup$
– Matti P.
Jan 16 at 11:46
$begingroup$
$f(x)$ would be maximum when $1/f(x)$ would be minimum.
$endgroup$
– Paras Khosla
Jan 16 at 14:11
$begingroup$
What are $a$ and $b$ on the second page?
$endgroup$
– Matti P.
Jan 16 at 11:43
$begingroup$
What are $a$ and $b$ on the second page?
$endgroup$
– Matti P.
Jan 16 at 11:43
$begingroup$
Hint: the derivative can be written as $$ f'(x) = frac{-2x(x^2 +1)(x^2-x-1)(x^4 + x^3 - x^2 - x -1)}{(x^6 + 2x^3 -1)^2} $$ Source: wolframalpha.com/input/…
$endgroup$
– Matti P.
Jan 16 at 11:46
$begingroup$
Hint: the derivative can be written as $$ f'(x) = frac{-2x(x^2 +1)(x^2-x-1)(x^4 + x^3 - x^2 - x -1)}{(x^6 + 2x^3 -1)^2} $$ Source: wolframalpha.com/input/…
$endgroup$
– Matti P.
Jan 16 at 11:46
$begingroup$
$f(x)$ would be maximum when $1/f(x)$ would be minimum.
$endgroup$
– Paras Khosla
Jan 16 at 14:11
$begingroup$
$f(x)$ would be maximum when $1/f(x)$ would be minimum.
$endgroup$
– Paras Khosla
Jan 16 at 14:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $x-frac{1}{x}=t$.
Thus, $t>0$ and by AM-GM we obtain:
$$frac{x^4-x^2}{x^6+2x^3-1}=frac{x-frac{1}{x}}{x^2-frac{1}{x^3}+2}=frac{t}{t^3+3t+2}=$$
$$=frac{1}{t^2+frac{2}{t}+3}leqfrac{1}{3sqrt[3]{t^2left(frac{1}{t}right)^2}+3}=frac{1}{6}.$$
The equality occurs for $t=1$, which says that we got a maximal value.
answered Jan 16 at 13:49
Michael RozenbergMichael Rozenberg
109k1896201
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add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Let $x-frac{1}{x}=t$.
Thus, $t>0$ and by AM-GM we obtain:
$$frac{x^4-x^2}{x^6+2x^3-1}=frac{x-frac{1}{x}}{x^2-frac{1}{x^3}+2}=frac{t}{t^3+3t+2}=$$
$$=frac{1}{t^2+frac{2}{t}+3}leqfrac{1}{3sqrt[3]{t^2left(frac{1}{t}right)^2}+3}=frac{1}{6}.$$
The equality occurs for $t=1$, which says that we got a maximal value.
answered Jan 16 at 13:49
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
Let $x-frac{1}{x}=t$.
Thus, $t>0$ and by AM-GM we obtain:
$$frac{x^4-x^2}{x^6+2x^3-1}=frac{x-frac{1}{x}}{x^2-frac{1}{x^3}+2}=frac{t}{t^3+3t+2}=$$
$$=frac{1}{t^2+frac{2}{t}+3}leqfrac{1}{3sqrt[3]{t^2left(frac{1}{t}right)^2}+3}=frac{1}{6}.$$
The equality occurs for $t=1$, which says that we got a maximal value.
answered Jan 16 at 13:49
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
Let $x-frac{1}{x}=t$.
Thus, $t>0$ and by AM-GM we obtain:
$$frac{x^4-x^2}{x^6+2x^3-1}=frac{x-frac{1}{x}}{x^2-frac{1}{x^3}+2}=frac{t}{t^3+3t+2}=$$
$$=frac{1}{t^2+frac{2}{t}+3}leqfrac{1}{3sqrt[3]{t^2left(frac{1}{t}right)^2}+3}=frac{1}{6}.$$
The equality occurs for $t=1$, which says that we got a maximal value.
answered Jan 16 at 13:49
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
Let $x-frac{1}{x}=t$.
Thus, $t>0$ and by AM-GM we obtain:
$$frac{x^4-x^2}{x^6+2x^3-1}=frac{x-frac{1}{x}}{x^2-frac{1}{x^3}+2}=frac{t}{t^3+3t+2}=$$
$$=frac{1}{t^2+frac{2}{t}+3}leqfrac{1}{3sqrt[3]{t^2left(frac{1}{t}right)^2}+3}=frac{1}{6}.$$
The equality occurs for $t=1$, which says that we got a maximal value.
answered Jan 16 at 13:49
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 16 at 13:49
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 16 at 13:49
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 16 at 13:49
Michael RozenbergMichael Rozenberg
109k1896201
109k1896201
add a comment |
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$begingroup$
What are $a$ and $b$ on the second page?
$endgroup$
– Matti P.
Jan 16 at 11:43
$begingroup$
Hint: the derivative can be written as $$ f'(x) = frac{-2x(x^2 +1)(x^2-x-1)(x^4 + x^3 - x^2 - x -1)}{(x^6 + 2x^3 -1)^2} $$ Source: wolframalpha.com/input/…
$endgroup$
– Matti P.
Jan 16 at 11:46
$begingroup$
$f(x)$ would be maximum when $1/f(x)$ would be minimum.
$endgroup$
– Paras Khosla
Jan 16 at 14:11