What is the extreme f(x)=3x^2-4x+1,0<= x<= 2 ?

The extreme f(x)=3x^2-4x+1,0<= x<= 2 is Maximum(0,1),Minimum(2/3 ,-1/3),Maximum(2,5)

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extreme f(x)=3x^2-4x+1,0<= x<= 2

extreme points

f (x) = 3 x^{2} - 4 x + 1, 0 \leq x \leq 2

Maximum (0, 1), Minimum (\frac{2}{3}, - \frac{1}{3}), Maximum (2, 5)

Solution steps

Find the critical points:

x = 0, x = \frac{2}{3}, x = 2

Domain of

3 x^{2} - 4 x + 1, 0 \leq x \leq 2 : 0 \leq x \leq 2

Find intervals:

Decreasing

: 0 < x < \frac{2}{3},

Increasing

: \frac{2}{3} < x < 2

Plug the extreme point

x = 0

into

3 x^{2} - 4 x + 1 \Rightarrow y = 1

Maximum (0, 1)

Plug the extreme point

x = \frac{2}{3}

into

3 x^{2} - 4 x + 1 \Rightarrow y = - \frac{1}{3}

Minimum (\frac{2}{3}, - \frac{1}{3})

Plug the extreme point

x = 2

into

3 x^{2} - 4 x + 1 \Rightarrow y = 5

Maximum (2, 5)

Maximum (0, 1), Minimum (\frac{2}{3}, - \frac{1}{3}), Maximum (2, 5)

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

f (x) = 5 + 2 x - 2 x^{2}

f (x y) = 2 x y - x^{2} - 2 y^{2} + 3 x + 4

f (x) = 2 x^{3} + 3 x^{2} - 36 x + 17

f (x) = x^{3} + 2 x^{2} - 7 x [0.2]

What is the extreme f(x)=3x^2-4x+1,0<= x<= 2 ?
The extreme f(x)=3x^2-4x+1,0<= x<= 2 is Maximum(0,1),Minimum(2/3 ,-1/3),Maximum(2,5)