What is the extreme f(x)=3x^2-24x+9,0<= x<= 9 ?

The extreme f(x)=3x^2-24x+9,0<= x<= 9 is Maximum(0,9),Minimum(4,-39),Maximum(9,36)

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extreme f(x)=3x^2-24x+9,0<= x<= 9

extreme points

f (x) = 3 x^{2} - 24 x + 9, 0 \leq x \leq 9

Maximum (0, 9), Minimum (4, - 39), Maximum (9, 36)

Solution steps

Find the critical points:

x = 0, x = 4, x = 9

Domain of

3 x^{2} - 24 x + 9, 0 \leq x \leq 9 : 0 \leq x \leq 9

Find intervals:

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < 9

Plug the extreme point

x = 0

into

3 x^{2} - 24 x + 9 \Rightarrow y = 9

Maximum (0, 9)

Plug the extreme point

x = 4

into

3 x^{2} - 24 x + 9 \Rightarrow y = - 39

Minimum (4, - 39)

Plug the extreme point

x = 9

into

3 x^{2} - 24 x + 9 \Rightarrow y = 36

Maximum (9, 36)

Maximum (0, 9), Minimum (4, - 39), Maximum (9, 36)

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

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

4 x + 3 y = 21

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

3 x + y = 3

What is the extreme f(x)=3x^2-24x+9,0<= x<= 9 ?
The extreme f(x)=3x^2-24x+9,0<= x<= 9 is Maximum(0,9),Minimum(4,-39),Maximum(9,36)