What is the extreme f(x)= 8/x+2pix^2 ?

The extreme f(x)= 8/x+2pix^2 is Minimum(\sqrt[3]{2/pi},4*2^{2/3}\sqrt[3]{pi}+2pi(2/pi)^{2/3})

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extreme f(x)= 8/x+2pix^2

extreme points

f (x) = \frac{8}{x} + 2 π x^{2}

Minimum (3 \frac{2}{π}, 4 \cdot 2^{\frac{2}{3}} 3 π + 2 π (\frac{2}{π})^{\frac{2}{3}})

Solution steps

f^{'} (x) = - \frac{8}{x ^{2}} + 4 π x

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{2}{3 4 3 π},

Increasing

: \frac{2}{3 4 3 π} < x < \infty

Plug

x = 3 \frac{2}{π}

into

\frac{8}{x} + 2 π x^{2} : 4 \cdot 2^{\frac{2}{3}} 3 π + 2 π (\frac{2}{π})^{\frac{2}{3}}

Minimum (3 \frac{2}{π}, 4 \cdot 2^{\frac{2}{3}} 3 π + 2 π (\frac{2}{π})^{\frac{2}{3}})

f (x, y) = 400 - 9 x^{2} - 36 y^{2}

f (x, y) = \frac{1}{3} x^{3} - x

f (x, y) = 4 - x^{4} + 2 x^{2} - y^{2}

f (x) = (x^{2} - 1)^{3}, - 1 \leq x \leq 6

f (x) = (x^{2} - 1)^{3}, - 1 \leq x \leq 4

What is the extreme f(x)= 8/x+2pix^2 ?
The extreme f(x)= 8/x+2pix^2 is Minimum(\sqrt[3]{2/pi},4*2^{2/3}\sqrt[3]{pi}+2pi(2/pi)^{2/3})