What is the extreme y=x^3-4x^2-3x+8 ?

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

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extreme y=x^3-4x^2-3x+8

extreme points

y = x^{3} - 4 x^{2} - 3 x + 8

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)

ln (t + 5)

y - 12 = 4 (x - 6)

f (x) = y^{2} = x + 25

f (x) = \frac{x - 4}{x - 2}

f (x) = \frac{9}{x ^{2} + 16} + \frac{1}{x ^{2} - 25}

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