What is the extreme f(x)=x^3-4x^2-3x^2+6 ?

The extreme f(x)=x^3-4x^2-3x^2+6 is Maximum(0,6),Minimum(14/3 ,-1210/27)

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extreme f(x)=x^3-4x^2-3x^2+6

extreme points

f (x) = x^{3} - 4 x^{2} - 3 x^{2} + 6

Maximum (0, 6), Minimum (\frac{14}{3}, - \frac{1210}{27})

Solution steps

f^{'} (x) = 3 x^{2} - 14 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{14}{3},

Increasing

: \frac{14}{3} < x < \infty

Plug

x = 0

into

x^{3} - 4 x^{2} - 3 x^{2} + 6 : 6

Maximum (0, 6)

Plug

x = \frac{14}{3}

into

x^{3} - 4 x^{2} - 3 x^{2} + 6 : - \frac{1210}{27}

Minimum (\frac{14}{3}, - \frac{1210}{27})

Maximum (0, 6), Minimum (\frac{14}{3}, - \frac{1210}{27})

f (x, y) = (x^{2} + y^{2}) (e^{x + y})

f (x) = x ln (4 x), (0, \infty)

- 8 x^{3} + 24 x + 9

f (x) = x^{3} - 3 x^{2} - 45 x + 25

f (x) = x^{3} - 3 x^{2} - 45 x + 22

What is the extreme f(x)=x^3-4x^2-3x^2+6 ?
The extreme f(x)=x^3-4x^2-3x^2+6 is Maximum(0,6),Minimum(14/3 ,-1210/27)