What is the extreme ln^2(x^2+x+1) ?

The extreme ln^2(x^2+x+1) is Minimum(-1,0),Maximum(-1/2 ,ln^2(3/4)),Minimum(0,0)

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extreme ln^2(x^2+x+1)

extreme points

ln^{2} (x^{2} + x + 1)

Minimum (- 1, 0), Maximum (- \frac{1}{2}, ln^{2} (\frac{3}{4})), Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{2 ln ( x ^{2} + x + 1 ) ( 2 x + 1 )}{x ^{2} + x + 1}

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

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

Decreasing

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

Increasing

: 0 < x < \infty

Plug

x = - 1

into

ln^{2} (x^{2} + x + 1) : 0

Minimum (- 1, 0)

Plug

x = - \frac{1}{2}

into

ln^{2} (x^{2} + x + 1) : ln^{2} (\frac{3}{4})

Maximum (- \frac{1}{2}, ln^{2} (\frac{3}{4}))

Plug

x = 0

into

ln^{2} (x^{2} + x + 1) : 0

Minimum (0, 0)

Minimum (- 1, 0), Maximum (- \frac{1}{2}, ln^{2} (\frac{3}{4})), Minimum (0, 0)

f (x) = - x^{3} + 12 x - 11

f (x, y) = (x - y) (36 - x y)

f (x) = x^{\frac{8}{3}}

f (x) = x^{2} - 9 y x^{2} - 7 x + 12

f (x) = 2 x^{3} - 9 x^{2} + 12 x + 6

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