What is the extreme f(x)=2x^2+(2048)/x+13 ?

The extreme f(x)=2x^2+(2048)/x+13 is Minimum(8,397)

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extreme f(x)=2x^2+(2048)/x+13

extreme points

f (x) = 2 x^{2} + \frac{2048}{x} + 13

Minimum (8, 397)

Solution steps

f^{'} (x) = 4 x - \frac{2048}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = 8

into

2 x^{2} + \frac{2048}{x} + 13 : 397

Minimum (8, 397)

f (x) = (x + 3)^{\frac{4}{3}}, - 6 \leq x \leq 6

x y - 2 x - 2 y - x^{2} - y^{2}

f (x) = 25 x - 25 x e^{- y} - 50 y - x^{2}

f (x) = csc^{2} (x) - \frac{2}{3} cot (x)

f (x) = \frac{3 x}{4 x ^{2} + 1}

What is the extreme f(x)=2x^2+(2048)/x+13 ?
The extreme f(x)=2x^2+(2048)/x+13 is Minimum(8,397)