What is the extreme f(x)=1+5/x-4/(x^2) ?

The extreme f(x)=1+5/x-4/(x^2) is Maximum(8/5 , 41/16)

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extreme f(x)=1+5/x-4/(x^2)

extreme points

f (x) = 1 + \frac{5}{x} - \frac{4}{x ^{2}}

Maximum (\frac{8}{5}, \frac{41}{16})

Solution steps

f^{'} (x) = \frac{8}{x ^{3}} - \frac{5}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{8}{5},

Decreasing

: \frac{8}{5} < x < \infty

Plug

x = \frac{8}{5}

into

1 + \frac{5}{x} - \frac{4}{x ^{2}} : \frac{41}{16}

Maximum (\frac{8}{5}, \frac{41}{16})

f (x) = \frac{2}{3} x - 5

f (x, y) = x^{3} - x y + 3 y^{2}

f (x) = 4 x^{4} + y^{2} - 2 x y + 1

f (x) = x^{3} - 3 x^{2} - 24 x + 5

f (x, y) = 12 x y - x^{3} - 6 y^{2}

What is the extreme f(x)=1+5/x-4/(x^2) ?
The extreme f(x)=1+5/x-4/(x^2) is Maximum(8/5 , 41/16)