What is the extreme f(x)=x^3-5x^2-8x-4 ?

The extreme f(x)=x^3-5x^2-8x-4 is Maximum(-2/3 ,-32/27),Minimum(4,-52)

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extreme f(x)=x^3-5x^2-8x-4

extreme points

f (x) = x^{3} - 5 x^{2} - 8 x - 4

Maximum (- \frac{2}{3}, - \frac{32}{27}), Minimum (4, - 52)

Solution steps

f^{'} (x) = 3 x^{2} - 10 x - 8

Find intervals:

Increasing

: - \infty < x < - \frac{2}{3},

Decreasing

: - \frac{2}{3} < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - \frac{2}{3}

into

x^{3} - 5 x^{2} - 8 x - 4 : - \frac{32}{27}

Maximum (- \frac{2}{3}, - \frac{32}{27})

Plug

x = 4

into

x^{3} - 5 x^{2} - 8 x - 4 : - 52

Minimum (4, - 52)

Maximum (- \frac{2}{3}, - \frac{32}{27}), Minimum (4, - 52)

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

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

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

f (x) = \frac{- x}{x}

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

What is the extreme f(x)=x^3-5x^2-8x-4 ?
The extreme f(x)=x^3-5x^2-8x-4 is Maximum(-2/3 ,-32/27),Minimum(4,-52)