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

The extreme f(x)=x^3-4x^2-3x+8 is Maximum(-1/3 , 230/27),Minimum(3,-10)

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

extreme

Maximum (- \frac{1}{3}, \frac{230}{27}), Minimum (3, - 10)

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 + 8 : \frac{230}{27}

Maximum (- \frac{1}{3}, \frac{230}{27})

Plug

x = 3

into

x^{3} - 4 x^{2} - 3 x + 8 : - 10

Minimum (3, - 10)

Maximum (- \frac{1}{3}, \frac{230}{27}), Minimum (3, - 10)

What is the extreme f(x)=x^3-4x^2-3x+8 ?
The extreme f(x)=x^3-4x^2-3x+8 is Maximum(-1/3 , 230/27),Minimum(3,-10)