What is the extreme f(x)=2x^3-9x^2+12x ?

The extreme f(x)=2x^3-9x^2+12x is Maximum(1,5),Minimum(2,4)

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extreme f(x)=2x^3-9x^2+12x

extreme points

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

Maximum (1, 5), Minimum (2, 4)

Solution steps

f^{'} (x) = 6 x^{2} - 18 x + 12

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 1

into

2 x^{3} - 9 x^{2} + 12 x : 5

Maximum (1, 5)

Plug

x = 2

into

2 x^{3} - 9 x^{2} + 12 x : 4

Minimum (2, 4)

Maximum (1, 5), Minimum (2, 4)

f (x) = 2 - cos (3 x)

4 x + 3 y = 21

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

3 x + y = 3

(- 1, 2), (2, - 4)

What is the extreme f(x)=2x^3-9x^2+12x ?
The extreme f(x)=2x^3-9x^2+12x is Maximum(1,5),Minimum(2,4)