What is the extreme f(x)=1+7/x-2/(x^2) ?

The extreme f(x)=1+7/x-2/(x^2) is Maximum(4/7 , 57/8)

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extreme f(x)=1+7/x-2/(x^2)

extreme points

f (x) = 1 + \frac{7}{x} - \frac{2}{x ^{2}}

Maximum (\frac{4}{7}, \frac{57}{8})

Solution steps

f^{'} (x) = \frac{4}{x ^{3}} - \frac{7}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{4}{7},

Decreasing

: \frac{4}{7} < x < \infty

Plug

x = \frac{4}{7}

into

1 + \frac{7}{x} - \frac{2}{x ^{2}} : \frac{57}{8}

Maximum (\frac{4}{7}, \frac{57}{8})

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

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

f (x) = x^{4} - 8 x^{2} + 8

f (x) = 6 x^{4} + 2 x^{3} - 12 x^{2} + 3

f (x, y) = x y + \frac{64}{x} + \frac{64}{y}

What is the extreme f(x)=1+7/x-2/(x^2) ?
The extreme f(x)=1+7/x-2/(x^2) is Maximum(4/7 , 57/8)