What is the extreme f(x)=x^3-6x^2+9x ?

The extreme f(x)=x^3-6x^2+9x is Maximum(1,4),Minimum(3,0)

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extreme f(x)=x^3-6x^2+9x

extreme points

f (x) = x^{3} - 6 x^{2} + 9 x

Maximum (1, 4), Minimum (3, 0)

Solution steps

f^{'} (x) = 3 x^{2} - 12 x + 9

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 1

into

x^{3} - 6 x^{2} + 9 x : 4

Maximum (1, 4)

Plug

x = 3

into

x^{3} - 6 x^{2} + 9 x : 0

Minimum (3, 0)

Maximum (1, 4), Minimum (3, 0)

\frac{1}{x ^{\frac{3}{2}} + 3 x}

tan (x) - 4

sin (t) - (cos (t) + sin (t))

3 cos (4 x)

f (x) = (2 x - 14)^{4}

What is the extreme f(x)=x^3-6x^2+9x ?
The extreme f(x)=x^3-6x^2+9x is Maximum(1,4),Minimum(3,0)