extreme f(x)=sqrt(82x^2+180x+100)

Lösung

extrempunkte

f (x) = 82 x^{2} + 180 x + 100

Minimum (- \frac{45}{41}, \frac{5 2}{41})

Schritte zur Lösung

f^{'} (x) = \frac{2 ( 41 x + 45 )}{82 x ^{2} + 180 x + 100}

Intervalle finden:

Absteigend

: - \infty < x < - \frac{45}{41},

Ansteigend

: - \frac{45}{41} < x < \infty

Stecke

x = - \frac{45}{41}

82 x^{2} + 180 x + 100 : \frac{5 2}{41}

Minimum (- \frac{45}{41}, \frac{5 2}{41})

e x t re m e P (x, y) = x^{2} + y^{2} + 16 x - 16 y

e x t re m e f (x) = - 0.001 x^{2} + 9 x - 1296

e x t re m e f (x) = \frac{( 500 + 500 \cdot x )}{1 + 0.04 \cdot x ^{2}}

e x t re m e \frac{( x ^{2} + 2 )}{( x ^{2} - 16 )}

e x t re m e y = 2 x^{2} + 5