What is the extreme f(x)=1+7/x-9/(x^2) ?

The extreme f(x)=1+7/x-9/(x^2) is Maximum(18/7 , 85/36)

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extreme f(x)=1+7/x-9/(x^2)

extreme points

f (x) = 1 + \frac{7}{x} - \frac{9}{x ^{2}}

Maximum (\frac{18}{7}, \frac{85}{36})

Solution steps

f^{'} (x) = \frac{18}{x ^{3}} - \frac{7}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{18}{7},

Decreasing

: \frac{18}{7} < x < \infty

Plug

x = \frac{18}{7}

into

1 + \frac{7}{x} - \frac{9}{x ^{2}} : \frac{85}{36}

Maximum (\frac{18}{7}, \frac{85}{36})

2 x^{2} - x^{4}

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

f (x) = Y (x) = x^{3} - 3 x^{2} - 9 x + 20

f (x) = - 3 x^{\frac{2}{3}}, - 1 \leq x \leq 1

f (x) = x^{3} + 3 x^{2} - 24 x, - 1 \leq x \leq 3

What is the extreme f(x)=1+7/x-9/(x^2) ?
The extreme f(x)=1+7/x-9/(x^2) is Maximum(18/7 , 85/36)