What is the extreme f(x)=-4x^3+12x^2+36+3 ?

The extreme f(x)=-4x^3+12x^2+36+3 is Minimum(0,39),Maximum(2,55)

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extreme f(x)=-4x^3+12x^2+36+3

extreme points

f (x) = - 4 x^{3} + 12 x^{2} + 36 + 3

Minimum (0, 39), Maximum (2, 55)

Solution steps

f^{'} (x) = - 12 x^{2} + 24 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

- 4 x^{3} + 12 x^{2} + 36 + 3 : 39

Minimum (0, 39)

Plug

x = 2

into

- 4 x^{3} + 12 x^{2} + 36 + 3 : 55

Maximum (2, 55)

Minimum (0, 39), Maximum (2, 55)

f (x) = 3^{\frac{2}{3}} - 2 x (- 1.1)

f (x) = - 0.001 x^{2} + 6 x - 1900

h (x) = 5 (x - 1)^{\frac{2}{3}}, 0 \leq x \leq 2

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

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

What is the extreme f(x)=-4x^3+12x^2+36+3 ?
The extreme f(x)=-4x^3+12x^2+36+3 is Minimum(0,39),Maximum(2,55)