What is the extreme f(x)=3+9x^2-6x^3 ?

The extreme f(x)=3+9x^2-6x^3 is Minimum(0,3),Maximum(1,6)

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extreme f(x)=3+9x^2-6x^3

extreme points

f (x) = 3 + 9 x^{2} - 6 x^{3}

Minimum (0, 3), Maximum (1, 6)

Solution steps

f^{'} (x) = - 18 x^{2} + 18 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 0

into

3 + 9 x^{2} - 6 x^{3} : 3

Minimum (0, 3)

Plug

x = 1

into

3 + 9 x^{2} - 6 x^{3} : 6

Maximum (1, 6)

Minimum (0, 3), Maximum (1, 6)

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

f (x, y) = (a - a^{2}) x + (4 a - 2 a^{2}) y

f (x) = x (x - 4)^{\frac{1}{3}}

f (x) = a r cos (1 - ∣ x ∣) - π

f (x, y) = 7 x y

What is the extreme f(x)=3+9x^2-6x^3 ?
The extreme f(x)=3+9x^2-6x^3 is Minimum(0,3),Maximum(1,6)