What is the extreme f(x)=x^3-8x^2-12x+1 ?

The extreme f(x)=x^3-8x^2-12x+1 is Maximum(-2/3 , 139/27),Minimum(6,-143)

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extreme f(x)=x^3-8x^2-12x+1

extreme points

f (x) = x^{3} - 8 x^{2} - 12 x + 1

Maximum (- \frac{2}{3}, \frac{139}{27}), Minimum (6, - 143)

Solution steps

f^{'} (x) = 3 x^{2} - 16 x - 12

Find intervals:

Increasing

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

Decreasing

: - \frac{2}{3} < x < 6,

Increasing

: 6 < x < \infty

Plug

x = - \frac{2}{3}

into

x^{3} - 8 x^{2} - 12 x + 1 : \frac{139}{27}

Maximum (- \frac{2}{3}, \frac{139}{27})

Plug

x = 6

into

x^{3} - 8 x^{2} - 12 x + 1 : - 143

Minimum (6, - 143)

Maximum (- \frac{2}{3}, \frac{139}{27}), Minimum (6, - 143)

f (x) = 3 cos (x), 0 \leq x \leq 4 π

4 x + 6 y - x^{2} - y^{2}

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

f (x) = 1 - x

f (x) = - 3 x^{2} - 5 x + 7

What is the extreme f(x)=x^3-8x^2-12x+1 ?
The extreme f(x)=x^3-8x^2-12x+1 is Maximum(-2/3 , 139/27),Minimum(6,-143)