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

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

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

extreme points

f (x) = 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) = 4 x^{7}

f (x) = x^{6} + \frac{x}{6}

y = - 57.17 x^{2} + 621.83 x + 1089

f (x) = x + \frac{( 1 )}{( x )}, 0.2 \leq x \leq 4

f (x, y) = ln (1 + 4 x^{2} + 6 y^{2})

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