What is the extreme f(x)=x^3-12x^2+36x+8 ?

The extreme f(x)=x^3-12x^2+36x+8 is Maximum(2,40),Minimum(6,8)

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extreme f(x)=x^3-12x^2+36x+8

extreme points

f (x) = x^{3} - 12 x^{2} + 36 x + 8

Maximum (2, 40), Minimum (6, 8)

Solution steps

f^{'} (x) = 3 x^{2} - 24 x + 36

Find intervals:

Increasing

: - \infty < x < 2,

Decreasing

: 2 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 2

into

x^{3} - 12 x^{2} + 36 x + 8 : 40

Maximum (2, 40)

Plug

x = 6

into

x^{3} - 12 x^{2} + 36 x + 8 : 8

Minimum (6, 8)

Maximum (2, 40), Minimum (6, 8)

y = x + 9

- 2 sin (- 4 x + \frac{π}{2})

f (x) = 16 + 3 x

(12.7) (- 2.11)

(5, 7), (3, 0)

What is the extreme f(x)=x^3-12x^2+36x+8 ?
The extreme f(x)=x^3-12x^2+36x+8 is Maximum(2,40),Minimum(6,8)