What is the extreme f(x)=2ln(x^2+3)-x ?

The extreme f(x)=2ln(x^2+3)-x is Minimum(1,4ln(2)-1),Maximum(3,2ln(12)-3)

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extreme f(x)=2ln(x^2+3)-x

extreme points

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

Minimum (1, 4 ln (2) - 1), Maximum (3, 2 ln (12) - 3)

Solution steps

f^{'} (x) = \frac{4 x}{x ^{2} + 3} - 1

Find intervals:

Decreasing

: - \infty < x < 1,

Increasing

: 1 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = 1

into

2 ln (x^{2} + 3) - x : 4 ln (2) - 1

Minimum (1, 4 ln (2) - 1)

Plug

x = 3

into

2 ln (x^{2} + 3) - x : 2 ln (12) - 3

Maximum (3, 2 ln (12) - 3)

Minimum (1, 4 ln (2) - 1), Maximum (3, 2 ln (12) - 3)

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

\frac{x ^{2}}{x ^{2} + 1}

\frac{4}{( x - 2 ) ^{3}}

f (x) = \frac{1}{2} \cdot 4^{x} + 1

2 y = 3 x + 5

What is the extreme f(x)=2ln(x^2+3)-x ?
The extreme f(x)=2ln(x^2+3)-x is Minimum(1,4ln(2)-1),Maximum(3,2ln(12)-3)