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

The extreme f(x)=18x+3x^2-4x^3 is Minimum(-1,-11),Maximum(3/2 , 81/4)

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

extreme points

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

Minimum (- 1, - 11), Maximum (\frac{3}{2}, \frac{81}{4})

Solution steps

f^{'} (x) = 18 + 6 x - 12 x^{2}

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < \frac{3}{2},

Decreasing

: \frac{3}{2} < x < \infty

Plug

x = - 1

into

18 x + 3 x^{2} - 4 x^{3} : - 11

Minimum (- 1, - 11)

Plug

x = \frac{3}{2}

into

18 x + 3 x^{2} - 4 x^{3} : \frac{81}{4}

Maximum (\frac{3}{2}, \frac{81}{4})

Minimum (- 1, - 11), Maximum (\frac{3}{2}, \frac{81}{4})

What is the extreme f(x)=18x+3x^2-4x^3 ?
The extreme f(x)=18x+3x^2-4x^3 is Minimum(-1,-11),Maximum(3/2 , 81/4)