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

The extreme f(x)=ln(2x^2-2x+3),0<= x<= 2 is Maximum(0,ln(3)),Minimum(1/2 ,ln(5/2)),Maximum(2,ln(7))

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

extreme

Maximum (0, ln (3)), Minimum (\frac{1}{2}, ln (\frac{5}{2})), Maximum (2, ln (7))

Solution steps

Find the critical points:

x = 0, x = \frac{1}{2}, x = 2

Domain of

ln (2 x^{2} - 2 x + 3), 0 \leq x \leq 2 : 0 \leq x \leq 2

Find intervals:

Decreasing

: 0 < x < \frac{1}{2},

Increasing

: \frac{1}{2} < x < 2

Plug the extreme point

x = 0

into

ln (2 x^{2} - 2 x + 3) \Rightarrow y = ln (3)

Maximum (0, ln (3))

Plug the extreme point

x = \frac{1}{2}

into

ln (2 x^{2} - 2 x + 3) \Rightarrow y = ln (\frac{5}{2})

Minimum (\frac{1}{2}, ln (\frac{5}{2}))

Plug the extreme point

x = 2

into

ln (2 x^{2} - 2 x + 3) \Rightarrow y = ln (7)

Maximum (2, ln (7))

Maximum (0, ln (3)), Minimum (\frac{1}{2}, ln (\frac{5}{2})), Maximum (2, ln (7))

What is the extreme f(x)=ln(2x^2-2x+3),0<= x<= 2 ?
The extreme f(x)=ln(2x^2-2x+3),0<= x<= 2 is Maximum(0,ln(3)),Minimum(1/2 ,ln(5/2)),Maximum(2,ln(7))