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

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

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

extreme points

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

Maximum (- 1, - 1), Minimum (2, 2 ln (2) + \frac{1}{2})

Solution steps

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

Find intervals:

Increasing

: - 2 < x < - 1,

Decreasing

: - 1 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 1

into

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

Maximum (- 1, - 1)

Plug

x = 2

into

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

Minimum (2, 2 ln (2) + \frac{1}{2})

Maximum (- 1, - 1), Minimum (2, 2 ln (2) + \frac{1}{2})

f (x) = 3 x^{2} + 5 x + 8

f (t) = t 4 - t^{2}

f (x) = y^{3} - x^{2} + 2 y^{2} - 4 x - 3

f (x, y) = 2 x^{3} + x y^{2} + 5 x^{2} + y^{2} + 3

\frac{x ^{2} - 48}{x - 7}

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