What is the extreme f(x)=x^3-4x^2-3x+7 ?

The extreme f(x)=x^3-4x^2-3x+7 is Maximum(-1/3 , 203/27),Minimum(3,-11)

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extreme f(x)=x^3-4x^2-3x+7

extreme points

f (x) = x^{3} - 4 x^{2} - 3 x + 7

Maximum (- \frac{1}{3}, \frac{203}{27}), Minimum (3, - 11)

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 + 7 : \frac{203}{27}

Maximum (- \frac{1}{3}, \frac{203}{27})

Plug

x = 3

into

x^{3} - 4 x^{2} - 3 x + 7 : - 11

Minimum (3, - 11)

Maximum (- \frac{1}{3}, \frac{203}{27}), Minimum (3, - 11)

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

f (x) = - 0.4 x^{2} + 90 x - 2000

f (x) = x^{3} + 3 x^{2} + 4, - 3 \leq x \leq 2

f (x) = \frac{( x - 5 ) ^{2}}{x + 9}

f (x) = x^{5} - 2 x^{3} + 1

What is the extreme f(x)=x^3-4x^2-3x+7 ?
The extreme f(x)=x^3-4x^2-3x+7 is Maximum(-1/3 , 203/27),Minimum(3,-11)