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

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

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

extreme points

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

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

Solution steps

f^{'} (x) = \frac{4 x - 2}{2 x ^{2} - 2 x + 3}

Find intervals:

Decreasing

: - \infty < x < \frac{1}{2},

Increasing

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

Plug

x = \frac{1}{2}

into

ln (2 x^{2} - 2 x + 3) : ln (\frac{5}{2})

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

x + 2 - x^{2}

f (x) = 12 x^{4} - 80 x^{3} - 24 x^{2} + 240 x + 720

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

f (x, y) = ln (x) + y^{3}

f (x) = 2 x^{3} - 33 x^{2} + 180 x + 6

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