What is the extreme y=ln(x+2)+4/(x+2) ?

The extreme y=ln(x+2)+4/(x+2) is Minimum(2,2ln(2)+1)

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extreme y=ln(x+2)+4/(x+2)

extreme

Minimum (2, 2 ln (2) + 1)

Solution steps

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

Find intervals:

Decreasing

: - 2 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 2

into

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

Minimum (2, 2 ln (2) + 1)

What is the extreme y=ln(x+2)+4/(x+2) ?
The extreme y=ln(x+2)+4/(x+2) is Minimum(2,2ln(2)+1)