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

The extreme f(x)=-4x^3+6x^2+6 is Minimum(0,6),Maximum(1,8)

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

extreme points

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

Minimum (0, 6), Maximum (1, 8)

Solution steps

f^{'} (x) = - 12 x^{2} + 12 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 0

into

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

Minimum (0, 6)

Plug

x = 1

into

- 4 x^{3} + 6 x^{2} + 6 : 8

Maximum (1, 8)

Minimum (0, 6), Maximum (1, 8)

f (x, y) = \frac{5}{2} x^{2} + x y^{2} + 5 y^{2}

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

y = - \frac{4 x}{x ^{2} + 1}

f (x) = \frac{- 2}{3} x^{3} - 6 x^{2} - 10 x + 80

What is the extreme f(x)=-4x^3+6x^2+6 ?
The extreme f(x)=-4x^3+6x^2+6 is Minimum(0,6),Maximum(1,8)