What is the extreme f(x)=x^3-6x^2+9x+4 ?

The extreme f(x)=x^3-6x^2+9x+4 is Maximum(1,8),Minimum(3,4)

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extreme f(x)=x^3-6x^2+9x+4

extreme points

f (x) = x^{3} - 6 x^{2} + 9 x + 4

Maximum (1, 8), Minimum (3, 4)

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 : 8

Maximum (1, 8)

Plug

x = 3

into

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

Minimum (3, 4)

Maximum (1, 8), Minimum (3, 4)

f (x) = \frac{( - 3 )}{( ∣ x + 2 ∣ )}

f (x) = \frac{x ^{2} - 1}{x ^{2} + 5 x + 6}

f (x) = \frac{5 x ^{2} + 4}{x ^{2} + 3 x - 10}

f (x) = \frac{2}{x + 4}

f (x) = x^{2} - 1

What is the extreme f(x)=x^3-6x^2+9x+4 ?
The extreme f(x)=x^3-6x^2+9x+4 is Maximum(1,8),Minimum(3,4)