AM-GM Inequality doubt

[Kanishk Sharma]

<p>For x >0,maximize f(x) = (1 +x)(1 +x)(1−x)Give your answer up to 3 decimal places.</p>

[Kanishk Sharma]

<p>can someone pls help...</p><p> </p>

[Saumik Karfa]

$1$ and $-1$ are the zeros of $f(x)$

Clearly $f(x) \to -\infty$ when $x\to \infty$

Now $f^{\prime}(x)=1-2x-3x^2$ and $f^{\prime\prime}(x)= -2-6x$

Therefore, $f^{\prime}(x)=0 \Rightarrow x= -1, -\frac 13 $

$f^{\prime\prime}(\frac 13)= -2-6(\frac 13)=-4<0$ [maxima]

Hence for $x>0$, $f(x)$ attains maximum value at $\frac 13$

Max Value is : $f(\frac 13)=\frac{32}{27}\approx 1.185$

[Kanishk Sharma]

<p>Thank you :)</p><p> </p>

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