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 points

y = ln (x + 2) + \frac{4}{x + 2}

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)

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

f (x) = 9 x - (5 x) ln (x), (0, \infty)

f (x, y) = x^{3} + y^{3} - 27 x y

f (x) = y^{5} - 3 x y

f (x, y) = x^{2} + y^{2} - 2 x + 5 y

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)