What is the extreme f(x)=3x^2-12x+7,0<= x<= 4 ?

The extreme f(x)=3x^2-12x+7,0<= x<= 4 is Maximum(0,7),Minimum(2,-5),Maximum(4,7)

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extreme f(x)=3x^2-12x+7,0<= x<= 4

extreme points

f (x) = 3 x^{2} - 12 x + 7, 0 \leq x \leq 4

Maximum (0, 7), Minimum (2, - 5), Maximum (4, 7)

Solution steps

Find the critical points:

x = 0, x = 2, x = 4

Domain of

3 x^{2} - 12 x + 7, 0 \leq x \leq 4 : 0 \leq x \leq 4

Find intervals:

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < 4

Plug the extreme point

x = 0

into

3 x^{2} - 12 x + 7 \Rightarrow y = 7

Maximum (0, 7)

Plug the extreme point

x = 2

into

3 x^{2} - 12 x + 7 \Rightarrow y = - 5

Minimum (2, - 5)

Plug the extreme point

x = 4

into

3 x^{2} - 12 x + 7 \Rightarrow y = 7

Maximum (4, 7)

Maximum (0, 7), Minimum (2, - 5), Maximum (4, 7)

(x - 1)^{3} - 3 x

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

f (x) = 40 x^{2} + \frac{600}{x}

(y^{2} - x^{2}) (y - 2)

U (x, y) = x y

What is the extreme f(x)=3x^2-12x+7,0<= x<= 4 ?
The extreme f(x)=3x^2-12x+7,0<= x<= 4 is Maximum(0,7),Minimum(2,-5),Maximum(4,7)