What is the extreme f(x)=x^3-6x^2+18 ?

The extreme f(x)=x^3-6x^2+18 is Maximum(0,18),Minimum(4,-14)

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extreme f(x)=x^3-6x^2+18

extreme points

f (x) = x^{3} - 6 x^{2} + 18

Maximum (0, 18), Minimum (4, - 14)

Solution steps

f^{'} (x) = 3 x^{2} - 12 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

x^{3} - 6 x^{2} + 18 : 18

Maximum (0, 18)

Plug

x = 4

into

x^{3} - 6 x^{2} + 18 : - 14

Minimum (4, - 14)

Maximum (0, 18), Minimum (4, - 14)

f (x) = 4 x + \frac{8}{x}

f (x) = - x^{2} + 229 x

f (x) = 6 x - 0.2 x^{2}, 0 \leq x \leq 30

5 x^{5} - 3 x^{4} + 2

f (x) = 4 - x, - 3 \leq x \leq 3

What is the extreme f(x)=x^3-6x^2+18 ?
The extreme f(x)=x^3-6x^2+18 is Maximum(0,18),Minimum(4,-14)