What is the extreme f(x)=x^3-4x^2-3x+9 ?

The extreme f(x)=x^3-4x^2-3x+9 is Maximum(-1/3 , 257/27),Minimum(3,-9)

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extreme f(x)=x^3-4x^2-3x+9

extreme points

f (x) = x^{3} - 4 x^{2} - 3 x + 9

Maximum (- \frac{1}{3}, \frac{257}{27}), Minimum (3, - 9)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - \frac{1}{3},

Decreasing

: - \frac{1}{3} < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - \frac{1}{3}

into

x^{3} - 4 x^{2} - 3 x + 9 : \frac{257}{27}

Maximum (- \frac{1}{3}, \frac{257}{27})

Plug

x = 3

into

x^{3} - 4 x^{2} - 3 x + 9 : - 9

Minimum (3, - 9)

Maximum (- \frac{1}{3}, \frac{257}{27}), Minimum (3, - 9)

cos (2 x + 5)

(4, 1), (- 2, - 1)

f (x) = x + \frac{1}{x}

f (x) = - 3 (a - 1)

f (x) = - x + 3

What is the extreme f(x)=x^3-4x^2-3x+9 ?
The extreme f(x)=x^3-4x^2-3x+9 is Maximum(-1/3 , 257/27),Minimum(3,-9)