extreme f(x)=(x^2+7x+13)/(x+4)

Lösung

extrempunkte

f (x) = \frac{x ^{2} + 7 x + 13}{x + 4}

Maximum (- 5, - 3), Minimum (- 3, 1)

Schritte zur Lösung

f^{'} (x) = \frac{x ^{2} + 8 x + 15}{( x + 4 ) ^{2}}

Intervalle finden:

Ansteigend

: - \infty < x < - 5,

Absteigend

: - 5 < x < - 4,

Absteigend

: - 4 < x < - 3,

Ansteigend

: - 3 < x < \infty

Stecke

x = - 5

\frac{x ^{2} + 7 x + 13}{x + 4} : - 3

Maximum (- 5, - 3)

Stecke

x = - 3

\frac{x ^{2} + 7 x + 13}{x + 4} : 1

Minimum (- 3, 1)

Maximum (- 5, - 3), Minimum (- 3, 1)

e x t re m e \frac{( 5 x - 8 )}{x}

e x t re m e f (x) = sin (x), - 5 \leq x \leq 6

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

e x t re m e f (x) = 8 + (9 + 5 x)^{\frac{2}{5}}

e x t re m e f (x, y) = 7 x^{2} + 8 y^{2} + 4 x - 5 y + 8