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

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})

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)