What is the extreme f(x)=1+5/x-8/(x^2) ?

The extreme f(x)=1+5/x-8/(x^2) is Maximum(16/5 , 57/32)

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extreme f(x)=1+5/x-8/(x^2)

extreme points

f (x) = 1 + \frac{5}{x} - \frac{8}{x ^{2}}

Maximum (\frac{16}{5}, \frac{57}{32})

Solution steps

f^{'} (x) = \frac{16}{x ^{3}} - \frac{5}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{16}{5},

Decreasing

: \frac{16}{5} < x < \infty

Plug

x = \frac{16}{5}

into

1 + \frac{5}{x} - \frac{8}{x ^{2}} : \frac{57}{32}

Maximum (\frac{16}{5}, \frac{57}{32})

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

y = 3 x - x^{3}

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

f (x) = \frac{x ^{2} - 4}{x ^{2} + 4}, - 4 \leq x \leq 4

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

What is the extreme f(x)=1+5/x-8/(x^2) ?
The extreme f(x)=1+5/x-8/(x^2) is Maximum(16/5 , 57/32)