What is the extreme f(x)=2x^2+(28)/x ?

The extreme f(x)=2x^2+(28)/x is Minimum(\sqrt[3]{7},6*7^{2/3})

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extreme f(x)=2x^2+(28)/x

extreme points

f (x) = 2 x^{2} + \frac{28}{x}

Minimum (37, 6 \cdot 7^{\frac{2}{3}})

Solution steps

f^{'} (x) = 4 x - \frac{28}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 37,

Increasing

: 37 < x < \infty

Plug

x = 37

into

2 x^{2} + \frac{28}{x} : 6 \cdot 7^{\frac{2}{3}}

Minimum (37, 6 \cdot 7^{\frac{2}{3}})

f (x) = ln (x^{2} + 3 x + 7), - 2 \leq x \leq 2

g (x) = x^{3} - x^{2} - x + 3

f (x) = - x^{2} + 10 x - 21, 3 \leq x \leq 7

f (x) = 4 x^{5} - 25 x^{4} - 40 x^{3} - 4

y = x 5 - x

What is the extreme f(x)=2x^2+(28)/x ?
The extreme f(x)=2x^2+(28)/x is Minimum(\sqrt[3]{7},6*7^{2/3})