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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Increasing

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

Decreasing

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

Increasing

: 0 < x < \infty

Plug

x = \frac{1}{e ^{2}} - 1

into

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

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

Plug

x = 0

into

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

Minimum (0, 0)

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

f (x) = - 2 x^{2} + 140 x

f (x) = x^{3} - 27 x, 0 \leq x \leq 6

f (x) = x^{3} - 27 x, 0 \leq x \leq 4

f (x) = x^{2} - 6 x - 1, 0 \leq x \leq 3

f (x) = x^{2} - 6 x - 1, 0 \leq x \leq 5

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