What is the extreme f(x)=x^3-5x^2-8x+9 ?

The extreme f(x)=x^3-5x^2-8x+9 is Maximum(-2/3 , 319/27),Minimum(4,-39)

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extreme f(x)=x^3-5x^2-8x+9

extreme points

f (x) = x^{3} - 5 x^{2} - 8 x + 9

Maximum (- \frac{2}{3}, \frac{319}{27}), Minimum (4, - 39)

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 + 9 : \frac{319}{27}

Maximum (- \frac{2}{3}, \frac{319}{27})

Plug

x = 4

into

x^{3} - 5 x^{2} - 8 x + 9 : - 39

Minimum (4, - 39)

Maximum (- \frac{2}{3}, \frac{319}{27}), Minimum (4, - 39)

f (x) = 5 x + 40

y^{2} = x^{2} + 9

f (x) = 2 - x

ln (x - 2)

y = x^{2} + 2

What is the extreme f(x)=x^3-5x^2-8x+9 ?
The extreme f(x)=x^3-5x^2-8x+9 is Maximum(-2/3 , 319/27),Minimum(4,-39)