What is the extreme f(x)=-4x^3+3x^2+18 ?

The extreme f(x)=-4x^3+3x^2+18 is Minimum(0,18),Maximum(1/2 , 73/4)

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extreme f(x)=-4x^3+3x^2+18

extreme points

f (x) = - 4 x^{3} + 3 x^{2} + 18

Minimum (0, 18), Maximum (\frac{1}{2}, \frac{73}{4})

Solution steps

f^{'} (x) = - 12 x^{2} + 6 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{1}{2},

Decreasing

: \frac{1}{2} < x < \infty

Plug

x = 0

into

- 4 x^{3} + 3 x^{2} + 18 : 18

Minimum (0, 18)

Plug

x = \frac{1}{2}

into

- 4 x^{3} + 3 x^{2} + 18 : \frac{73}{4}

Maximum (\frac{1}{2}, \frac{73}{4})

Minimum (0, 18), Maximum (\frac{1}{2}, \frac{73}{4})

f (x) = 2 x^{\frac{2}{3}}, - 27 \leq x \leq 27

f (x) = (x^{2} - 1)^{3}, - 1 \leq x \leq 5

f (x) = \frac{1}{3} x^{3} - 9 x^{2} + 72 x + 2

F (x, y) = x^{3} y^{2} + x^{2} y + 2 x y^{2} + 2 y

f (x) = \frac{1}{x} + x

What is the extreme f(x)=-4x^3+3x^2+18 ?
The extreme f(x)=-4x^3+3x^2+18 is Minimum(0,18),Maximum(1/2 , 73/4)