3 years ago

What is the solution to x -2<10

Mathematics

1 answer:

andriy [413]3 years ago

6 0

There ya gooooooooooooon

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Xe^(-x^2/128) absolute max and absolute min

adoni [48]

$f(x)=xe^{-x^2/128}$
$\implies f'(x)=e^{-x^2/128}+x\left(-\dfrac{2x}{128}\right)e^{-x^2/128}=\left(1-\dfrac1{64}x^2\right)e^{-x^2/128}$

Extrema can occur when the derivative is zero or undefined.

$\left(1-\dfrac1{64}x^2\right)e^{-x^2/128}=0\implies 1-\dfrac1{64}x^2=0\implies x^2=64\implies x=\pm8$

Maxima occur where the first derivative is zero and the second derivative is negative; minima where the second derivative is positive. You have

$f''(x)=-\dfrac1{32}xe^{-x^2/128}+\left(1-\dfrac1{64}x^2\right)\left(-\dfrac{2x}{128}\right)e^{-x^2/128}=\left(-\dfrac3{64}x+\dfrac1{4096}x^3\right)e^{-x^2/128}$

At the critical points, you get

$f''(-8)=\dfrac1{4\sqrt e}>0$
$f''(8)=-\dfrac1{4\sqrt e}$

So you have a minimum at $\left(-8,-\dfrac8{\sqrt e}\right)$ and a maximum at $\left(8,\dfrac8{\sqrt e}\right)$ .

Meanwhile, as $x\to\pm\infty$ , it's clear that $f(x)\to0$ , so these extrema are absolute on the function's domain.