What is the extreme f(x)=x^2+(240)/x ?

The extreme f(x)=x^2+(240)/x is Minimum(2\sqrt[3]{15},12*15^{2/3})

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extreme f(x)=x^2+(240)/x

extreme points

f (x) = x^{2} + \frac{240}{x}

Solution steps

f^{'} (x) = 2 x - \frac{240}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

Increasing

Plug

into

x^{2} + \frac{240}{x} : 12 \cdot 1 5^{\frac{2}{3}}

f (x) = 6 x^{4} - 8 x^{3} - 24 x^{2} + 35

f (x) = x + \frac{13}{x}

f (x) = \frac{2 x}{x ^{2} + 9}

2 x^{3} - 4 x^{2} + 2 x + 1

y (x, t) = x (3 t)

What is the extreme f(x)=x^2+(240)/x ?
The extreme f(x)=x^2+(240)/x is Minimum(2\sqrt[3]{15},12*15^{2/3})