What is the extreme f(x)=f(x)=x^3-4x^2-3x+1 ?

The extreme f(x)=f(x)=x^3-4x^2-3x+1 is Maximum(-1/3 , 41/27),Minimum(3,-17)

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extreme f(x)=f(x)=x^3-4x^2-3x+1

extreme points

f (x) = f (x) = x^{3} - 4 x^{2} - 3 x + 1

Maximum (- \frac{1}{3}, \frac{41}{27}), Minimum (3, - 17)

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 + 1 : \frac{41}{27}

Maximum (- \frac{1}{3}, \frac{41}{27})

Plug

x = 3

into

x^{3} - 4 x^{2} - 3 x + 1 : - 17

Minimum (3, - 17)

Maximum (- \frac{1}{3}, \frac{41}{27}), Minimum (3, - 17)

f (x) = x^{3} + 3 x^{2} - 189 x

f (x) = 1 - ∣ x - 7 ∣

f (x, y) = (x^{2} - 49)^{2} + (y^{2} - 9)^{2}

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

x^{3} - 3 x + 1, - \frac{3}{2} \leq x \leq 3

What is the extreme f(x)=f(x)=x^3-4x^2-3x+1 ?
The extreme f(x)=f(x)=x^3-4x^2-3x+1 is Maximum(-1/3 , 41/27),Minimum(3,-17)