What is the extreme f(x)=x^3-x^2-8x+12 ?

The extreme f(x)=x^3-x^2-8x+12 is Maximum(-4/3 , 500/27),Minimum(2,0)

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extreme f(x)=x^3-x^2-8x+12

extreme

Maximum (- \frac{4}{3}, \frac{500}{27}), Minimum (2, 0)

Solution steps

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

Find intervals:

Increasing

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

Decreasing

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

Increasing

: 2 < x < \infty

Plug

x = - \frac{4}{3}

into

x^{3} - x^{2} - 8 x + 12 : \frac{500}{27}

Maximum (- \frac{4}{3}, \frac{500}{27})

Plug

x = 2

into

x^{3} - x^{2} - 8 x + 12 : 0

Minimum (2, 0)

Maximum (- \frac{4}{3}, \frac{500}{27}), Minimum (2, 0)

What is the extreme f(x)=x^3-x^2-8x+12 ?
The extreme f(x)=x^3-x^2-8x+12 is Maximum(-4/3 , 500/27),Minimum(2,0)