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

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

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

extreme points

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

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

Solution steps

f^{'} (x) = 3 (x - 3) (x - 1)

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 1

into

x (x^{2} - 6 x + 9) : 4

Maximum (1, 4)

Plug

x = 3

into

x (x^{2} - 6 x + 9) : 0

Minimum (3, 0)

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

y = \frac{1 - x}{cos ( x )}

- sin (2 x)

f (x) = lo g_{2} (x)

f (x) = \frac{x ^{3}}{x ^{4} - 1}

4 - x^{2}

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