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

Lösung

extrempunkte

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)

Schritte zur Lösung

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

Intervalle finden:

Absteigend

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

Ansteigend

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

Absteigend

: 5 < x < \infty

Stecke

x = \frac{1}{5}

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

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

Stecke

x = 5

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)

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

in v erse \frac{π}{2} tan (x)

in v erse f (x) = 80 - 0.2 x

s l o p e in t erce pt 4 x + 3 y = - 6

s l o p e q = 20 - 2 p