What is the extreme f(x)=13ln(x^2+1)-5x ?

The extreme f(x)=13ln(x^2+1)-5x is Minimum(1/5 ,13ln(26/25)-1),Maximum(5,13ln(26)-25)

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extreme f(x)=13ln(x^2+1)-5x

extreme points

f (x) = 13 ln (x^{2} + 1) - 5 x

Minimum (\frac{1}{5}, 13 ln (\frac{26}{25}) - 1), Maximum (5, 13 ln (26) - 25)

Solution steps

f^{'} (x) = \frac{26 x}{x ^{2} + 1} - 5

Find intervals:

Decreasing

: - \infty < x < \frac{1}{5},

Increasing

: \frac{1}{5} < x < 5,

Decreasing

: 5 < x < \infty

Plug

x = \frac{1}{5}

into

13 ln (x^{2} + 1) - 5 x : 13 ln (\frac{26}{25}) - 1

Minimum (\frac{1}{5}, 13 ln (\frac{26}{25}) - 1)

Plug

x = 5

into

13 ln (x^{2} + 1) - 5 x : 13 ln (26) - 25

Maximum (5, 13 ln (26) - 25)

Minimum (\frac{1}{5}, 13 ln (\frac{26}{25}) - 1), Maximum (5, 13 ln (26) - 25)

ln (11 x^{2} + 3)

f (x) = \frac{x + 3}{x + 2}

f (x) = (x - 5)^{3} - 1

5 x + y = 4

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

What is the extreme f(x)=13ln(x^2+1)-5x ?
The extreme f(x)=13ln(x^2+1)-5x is Minimum(1/5 ,13ln(26/25)-1),Maximum(5,13ln(26)-25)