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

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

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extreme f(x)=3x^2-12x+5,0<= x<= 3

extreme points

f (x) = 3 x^{2} - 12 x + 5, 0 \leq x \leq 3

Maximum (0, 5), Minimum (2, - 7), Maximum (3, - 4)

Solution steps

Find the critical points:

x = 0, x = 2, x = 3

Domain of

3 x^{2} - 12 x + 5, 0 \leq x \leq 3 : 0 \leq x \leq 3

Find intervals:

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < 3

Plug the extreme point

x = 0

into

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

Maximum (0, 5)

Plug the extreme point

x = 2

into

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

Minimum (2, - 7)

Plug the extreme point

x = 3

into

3 x^{2} - 12 x + 5 \Rightarrow y = - 4

Maximum (3, - 4)

Maximum (0, 5), Minimum (2, - 7), Maximum (3, - 4)

f (x) = x - sin (x), 0 \leq x \leq 2 π

f (x) = 4 x^{2} + y^{2} - 4

y = x^{2} - 2 x - 8

f (x, y) = 5^{2} x^{4} + 5^{2} y^{4} - 4 x y + 14

f (x) = x^{3} - 3 x^{2} + x - 1

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