What is the extreme f(x)=((8x^3-x^2-108))/(4x^2) ?

The extreme f(x)=((8x^3-x^2-108))/(4x^2) is Maximum(-3,-37/4)

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extreme f(x)=((8x^3-x^2-108))/(4x^2)

extreme points

f (x) = \frac{( 8 x ^{3} - x ^{2} - 108 )}{4 x ^{2}}

Maximum (- 3, - \frac{37}{4})

Solution steps

f^{'} (x) = \frac{2 ( x ^{3} + 27 )}{x ^{3}}

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 3

into

\frac{8 x ^{3} - x ^{2} - 108}{4 x ^{2}} : - \frac{37}{4}

Maximum (- 3, - \frac{37}{4})

f (x) = 3 x - 4

- (x + 1)^{2} - 5 (y - 3)^{2}

f (x) = ln (x + y + 1)

f (x) = 3 - x^{3}

f (x) = [0.1]

What is the extreme f(x)=((8x^3-x^2-108))/(4x^2) ?
The extreme f(x)=((8x^3-x^2-108))/(4x^2) is Maximum(-3,-37/4)