What is the extreme f(x)=x^3-18x^2+96x ?

The extreme f(x)=x^3-18x^2+96x is Maximum(4,160),Minimum(8,128)

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extreme f(x)=x^3-18x^2+96x

extreme points

f (x) = x^{3} - 18 x^{2} + 96 x

Maximum (4, 160), Minimum (8, 128)

Solution steps

f^{'} (x) = 3 x^{2} - 36 x + 96

Find intervals:

Increasing

: - \infty < x < 4,

Decreasing

: 4 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = 4

into

x^{3} - 18 x^{2} + 96 x : 160

Maximum (4, 160)

Plug

x = 8

into

x^{3} - 18 x^{2} + 96 x : 128

Minimum (8, 128)

Maximum (4, 160), Minimum (8, 128)

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

f (x, y) = 3 x^{2} + 2 y^{2} - x y - 4 x - 7 y + 12

\frac{x ^{3} - 4}{x ^{2}}

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

f (x) = 5 - 7 x^{2}

What is the extreme f(x)=x^3-18x^2+96x ?
The extreme f(x)=x^3-18x^2+96x is Maximum(4,160),Minimum(8,128)